A compound, C11H12O2, has an IR spectrum showing a peak at 1710 cm-1. Its 1H NMR spectrum has peaks at delta 1.3 (3 H, triplet), 4.3 (2 H, quartet), 6.5 (1 H, doublet), 7.4-7.6 (5 H, multiplet), and 7.7 (1 H, doublet).

Required:
Draw its structure below.

Answer:

3.97 L

Explanation:

Data obtained from the question include the following:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 19 °C

Final temperature (T2) = 58 °C

Final volume (V2) =..?

Next, we shall convert celsius temperature to Kelvin temperature. This can be done as shown below:

Temperature (K) = temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273 = 292 K

Final temperature (T2) = 58 °C

Final temperature (T2) = 58 °C + 273 = 331 K

Finally, we shall determine the new volume of the gas by using Charles' law equation as shown below:

Initial volume (V1) = 3.5 L

Initial temperature (T1) = 292 K

Final temperature (T2) = 331 K

Final volume (V2) =..?

V1 /T1 = V2 /T2

3.5 /292 = V2 /331

Cross multiply

292 x V2 = 3.5 x 331

Divide both side by 292

V2 = (3.5 x 331) / 292

V2 = 3.97 L

Therefore, the new volume of the gas is 3.97 L.

Answer:

Polar molecules are not symmetrical

Explanation:

Even though the structures of the molecules involved were not shown in the question, but I will proceed to give a general explanation of the conditions that describe a polar molecule.

First of all, symmetrical molecules are non-polar and asymmetrical molecules are polar. This is the reason why CF4 will be a nonpolar molecule but H2O will be a polar molecule. Some symmetrical molecules may may posses polar bonds or dipoles but these dipoles eventually cancel out since the molecule is symmetrical in nature.

Summarily, if a molecule possess the same type of atoms attached to the central atom with some symmetry axes, like C3, C4 etc., we will end up with a non polar molecule but if we have a nonplanar molecule, then we will end up finding it to be polar.

Answer:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

Answer:

760 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

Answer:

Salt solution may be calculated by measuring the specific gravity of a sample of water using a hydrometer.

Hope this answer correct (^^)....

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

Answer:

A.

Explanation:

Did it on Edge.

The given question is incomplete, the complete question is:

A chemistry student weighs out 0.0349g of formic acid HCHO2 into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1500M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

Answer:

The correct answer is 5.06 ml.

Explanation:

Based on the given information, the weight of formic acid given is 0.0349 grams. The volume of formic acid of V1 given is 250 ml. The molecular mass of formic acid is 46 grams per mole. Now the molarity of formic acid will be,  

[HCOOH] = weight * 1000 / molecular mass * volume (ml)

= 0.0349 * 1000 / 46 * 250

= 0.003035 M or M1

The molarity of NaOH given is 0.1500 M or M2

Let us assume that the volume needed to attain equivalence point is V2 ml.  The volume V2 can be determined by using the dilution equation,  

M1V1 = M2V2

V2 = M1V1/M2  

V2 = 0.003035 * 250 / 0.1500

V2 = 5.06 ml.  

Hence, the volume of NaOH needed is 5.06 ml.  

Answer:

[tex]m_{Si}=37.2gSi[/tex]

Explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]

We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:

[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]

Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:

[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]

Best regards.

Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

Learn more: https://brainly.com/question/3902440

Answer:

THE FINAL PRESSURE OF AMMONIA IF THERE IS NO CHANGE IN TEMPERATURE AND A DECREASE IN VOLUME FROM 90.3 mL TO 43.4 mL IS 2.91 atm.

Explanation:

At constant temperature, the pressure of a given mass of gas is inversely proportional to the volume. This question follows Boyle's law of gas laws.

Mathematically written as:

P1V1 = P2V2

Re-arranging the formula by making P2 the subject of the formula;

P2 = P1V1 / T2

P1 = 1.4 atm

V1 = 90.3 mL

V2 = 43.4 mL

P2 = unknown

So therefore, we have:

P2 = 1.4 * 90.3 / 43.4

P2 = 2.91 atm

The final pressure of ammonia is therefore 2.91 atm.

Answer:

2.37 atm

Explanation:

Step 1: Given data

Step 2: Calculate the final pressure of ammonia

Since the temperature is kept constant, we can calculate the final pressure of ammonia using Boyle's law.

[tex]P_1 \times V_1 = P_2 \times V_2\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{1.14atm \times 90.3mL}{43.4mL} = 2.37 atm[/tex]

Answer:

See attached picture.

Explanation:

Hello,

In this case, starting by 1-butanol, we can make it react with hydrogen bromide (1st step) in order to yield 1-bromobutane. Next, the formed alkyl halide is treated magnesium in the presence of an ether in order to yield butyl magnesium bromide which is a Grignard reagent (2nd step). Finally, by adding carbon dioxide, water and extra hydrogen chloride, a carbonyl group can be formed between two butyl radicals in order to form the 5-nonanone (3rd step) as shown on the attached picture.

Best regards.

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

0.60 mol

Explanation:

There is some info missing. I think this is the original question.

Acetylene gas is often used in welding torches because of the very high heat produced when it reacts with oxygen gas, producing carbon dioxide gas and water vapor. Calculate the moles of oxygen needed to produce 1.5 mol of water.

Step 1: Given data

Moles of water required: 1.5 mol

Step 2: Write the balanced equation

C₂H₂(g) + 2.5 O₂(g) ⇒ 2 CO₂(g) + H₂O(g)

Step 3: Calculate the moles of oxygen needed to produce 1.5 mol of water

The molar ratio of O₂ to H₂O is 2.5:1. The moles of oxygen needed to produce 1.5 mol of water are (1/2.5) × 1.5 mol = 0.60 mol

Answer:

b, H2SO4 with HCl, as they are both liquid acids

Answer:

hi here goes your answer

Explanation:

iv. The lower the PH, the weaker the base

Answer:

pH of the buffer is 7.48

Explanation:

The H₂PO₄⁻/HPO₄²⁻ buffer has a pKa of 7.21. You can find pH of this buffer following H-H equation:

pH = pKa + log [A⁻] / [HA]

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

Where [] represents molarity of each specie of the buffer and, as volume is 1.00L, also represents its moles.

Thus, to find pH of the buffer we need to calculate moles of each specie, thus

Moles of 18.0g of KH₂PO₄(Molar mass: 136.086g/mol) = moles of H₂PO₄⁻ are:

18.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.132 moles of KH₂PO₄= H₂PO₄⁻

Moles of 35.0g of Na₂HPO₄(Molar mass: 141.96g/mol) = moles of HPO₄²⁻ are:

35.0g Na₂HPO₄ ₓ (1mol / 141.96g) = 0.2465 moles of Na₂HPO₄= HPO₄²⁻

Replacing in H-H equation:

pH = 7.21 + log [HPO₄²⁻] / [ H₂PO₄⁻]

pH = 7.21 + log [0.2465] / [0.132]

pH = 7.48

Answer:

1) 3-Ethylanisole

2) 2,3-Dibromobenzoic acid

3) 3-Fluorotoluene

Explanation:

Let us try to look at the structures of each compound one after the other as described in the question.

1) A ring with six vertices and alternating double and single bonds must refer to a benzene ring. A benzene ring having -OCH3 attached to the first vertex is called anisole. If a -CH2CH3 group is now attached at position 3, we now name the compound 3-Ethylanisole.

2) A ring with six vertices and alternating single and double bonds is a benzene ring. If the ring has -COOH attached to the first vertex, we call it benzoic acid. If bromine atoms are attached to the second and third vertices respectively, the compound is now named 2,3-Dibromobenzoic acid.

3) A ring with alternating single and double bonds is a benzene ring. If a -CH3 group is attached to the first vertex, we call the compound toluene. If a fluorine atom is now attached to position 3, the compound can now be named 3-Fluorotoluene

Answer:

See explanation

Explanation:

The calculated concentration of the sodium hydroxide is;

Number of moles= mass/molar mass = 1g/40gmol-1 = 0.025 moles

Concentration= number of moles/volume= 0.025×1000/250 = 0.1 M

This calculated concentration will be different from the molarity of NaOH obtained by standardization with acid. The result will not be more accurate if a different acid is used for the standardization this is because sodium hydroxide is deliquescent and absorbs moisture thereby leading to inaccuracy in the calculated molarity.

Any substance that must be used as a primary standard must not absorb moisture, it must be stable and it must be a substance in its pure form.

Answer:

Question 1

A. Empirical formula is C8H8O3

B. Molecular formula is C8H8O3

Question 2.

A. Empirical formula is CH2

B. Molecular formula is C4H8

Explanation:

Question 1:

A. Determination of the empirical formula:

Carbon (C) = 63.2%

Hydrogen (H) = 5.26%

Oxygen (O) = 31.6%

Divide by their molar mass

C = 63.2/12 = 5.27

H = 5.26/1 = 5.26

O = 31.6/16 = 1.975

Divide by the smallest

C = 5.27/1.975 = 2.7

H = 5.26/1.975 = 2.7

O = 1.975/1.975 = 1

Multiply through by 3 to express in whole number

C = 2.7 x 3 = 8

H = 2.7 x 3 = 8

O = 1 x 3 = 3

Therefore, the empirical formula for the compound is C8H8O3

B. Determination of the molecular formula of the compound.

From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules.

Now from the question given, we were told that 1 molecule of the compound has a mass of 2.53×10¯²² g.

Therefore, 6.02×10²³ molecules will have a mass of = 6.02×10²³ x 2.53×10¯²² = 152.306 g

Therefore, 1 mole of the compound = 152.306 g

The molecular formula of the compound can be obtained as follow:

[C8H8O3]n = 152.306

[(12x8) + (1x8) + (16x3)]n = 152.306

[(96 + 8 + 48 ]n = 152.306

152n = 152.306

Divide both side by 152

n = 152.306/152

n = 1

The molecular formula => [C8H8O3]n

=> [C8H8O3]1

=> C8H8O3

Question 2:

A. Determination of the empirical formula of the compound.

Mass sample of compound = 0.648 g

Carbon (C) = 0.556 g

Mass of Hydrogen (H) = mass sample of compound – mass of carbon

Mass of Hydrogen (H) = 0.648 – 0.556

Mass of Hydrogen (H) = 0.092 g

Thus, the empirical formula can be obtained as follow:

C = 0.556 g

H = 0.092 g

Divide by their molar mass

C = 0.556/12 = 0.046

H = 0.092/1 = 0.092

Divide by the smallest

C = 0.046/0.046 = 1

H = 0.092/0.046 = 2

Therefore, the empirical formula of the compound is CH2.

B. Determination of the molecular formula of the compound.

Mole of compound = 0.5 mole

Mass of compound = 28.5 g

Molar mass of compound =.?

Mole = mass /Molar mass

0.5 = 28.5/ Molar mass

Cross multiply

0.5 x molar mass = 28.5

Divide both side by 0.5

Molar mass = 28.5/0.5 = 57 g/mol

Thus, the molecular formula of compound can be obtained as follow:

[CH2]n = 57

[12 + (1x2)]n = 57

14n = 57

Divide both side by 14

n = 57/14

n = 4

Molecular formula => [CH2]n

=> [CH2]4

=> C4H8.

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

Learn more:

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Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

2. We'll begin by writing the balanced equation for the reaction. This is given below:

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

Answer:

The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Explanation:

The half reactions are:

2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)

Sn(s) ⟶ Sn²⁺(aq) + 2 e-  (oxidation)

In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:

anode reaction∥ cathode reaction

where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.  

In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).

The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:

Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

Answer:

Acid spills should be neutralized with sodium bicarbonate and then cleaned up with a paper towel or sponge.

Explanation:

Answer:

28.13 g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2O4(l) + 2N2H4 (l) → 3N2(g) + 4H2O(g)

Next, we shall determine the masses of N2O4 and N2H4 that reacted and the mass of H2O produced from the balanced equation.

This is illustrated below:

Molar mass of N2O4 = (14x2) + (16x4) = 92 g/mol

Mass of N2O4 from the balanced equation = 1 x 92 = 92g

Molar mass of N2H4 = (14x2) + (4x1) = 32 g/mol

Mass of N2H4 from the balanced equation = 2 x 32 = 64 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72 g

Summary:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4 to produce 72 g of H2O.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

92 g of N2O4 reacted with 64 g of N2H4.

Therefore, 40 g of N2O4 will react with = (40 x 64)/92 = 27.83 g of N2H4.

From the calculations made above, we can see that it will take a higher mass i.e 27.83 g than what was given i.e 25 g of N2H4 to react completely with 40 g of N2O4.

Therefore, N2H4 is the limiting reactant and N2O4 is the excess reactant.

Finally, we shall determine the mass of H2O produced from the reaction of 40.0 g of N2O4 and 25.0 g of N2H4.

In this case the limiting reactant will be used because it will produce the maximum amount of H2O as all of it is consumed in the reaction.

The limiting reactant is N2H4 and the mass of H2O produced can be obtained as follow:

From the balanced equation above,

64 g of N2H4 reacted to produce 72 g of H2O.

Therefore, 25 g of N2H4 will react to produce = (25 x 72)/64 = 28.13 g of H2O.

Therefore, 28.13 g of H2O were obtained from the reaction.