In this reaction, how many grams of O2 are required to completely react with 110 grams of Al

Answers

Answer 1

Answer:

98 g

Explanation:

Start with the balanced equation. Then, make a little chart under the equation showing the information you have and need.

        3O₂(g) + 4Al(s)  →  2Al₂O₃(s)

m        ?           110 g

M      ___       ____

n       ___ <=  ____

"m" is for mass. "M" is for molar mass (some teachers use "MM"). "n" is for the number of moles.

To find the mass of oxygen:

Calculate the molar mass of oxygen ([tex]M_{O_{2}[/tex])Calculate molar mass of aluminum ([tex]M_{Al}[/tex])Use [tex]M_{Al}[/tex] to find the moles of aluminum ([tex]n_{Al}[/tex])With [tex]n_{Al}[/tex], use the mole ratio to find the moles of oxygen ([tex]n_{O_{2}}[/tex])Use [tex]n_{O_{2}}[/tex] and [tex]M_{O_{2}}[/tex] to find the mass of oxygen ([tex]m_{O_{2}}[/tex])

To find molar mass, use the atomic mass on your periodic table. For each atom of an element, add on its atomic mass.

Molar mass of aluminum (one Al atom):

[tex]M_{Al} = 26.982 g/mol[/tex]

Molar mass of oxygen (two O atoms):

[tex]M_{O_{2}} = 16.000g/mol+16.000g/mol[/tex]

[tex]= 32.000g/mol[/tex]

Update the chart:

               3O₂(g)       +          4Al(s)          →     2Al₂O₃(s)

m                ?                        110 g

M      32.000 g/mol       26.982 g/mol

n                ___         <=      ____

Find the moles of aluminum

[tex]n_{Al} = \frac{110g}{1} *\frac{1mol}{26.982g}[/tex]    Multiply mass by molar mass to find moles.

[tex]n_{Al} = \frac{110}{1} *\frac{1mol}{26.982}[/tex]       The units "g" cancel out.

[tex]n_{Al} = 4.0(7)mol[/tex]        Keep one extra significant figure. (110 has 2 sig. figs.)

               3O₂(g)       +          4Al(s)          →     2Al₂O₃(s)

m                ?                        110 g

M      32.000 g/mol       26.982 g/mol

n                ___         <=   4.0(7) mol

Find the moles of oxygen using the mole ratio, which comes from the coefficients in the balanced equation.

The mole ratio of oxygen to aluminum is 3 to 4.

[tex]n_{O_{2}} = \frac{4.0(7)mol_{Al}}{1}*\frac{3mol_{O2}}{4mol_{Al}}[/tex]     Multiply moles of aluminum by the mole ratio.

[tex]n_{O_{2}} = \frac{4.0(7)}{1}*\frac{3mol_{O2}}{4}[/tex]             "molAl" units cancel out.

[tex]n_{O_{2}} = 3.0(52)mol_{O2}[/tex]              Keep two sig. figs. when the first is a "5"

               3O₂(g)       +          4Al(s)          →     2Al₂O₃(s)

m                ?                        110 g

M      32.000 g/mol       26.982 g/mol

n          3.0(52) mol   <=    4.0(7) mol

Find the mass of oxygen

[tex]m_{O_{2}} = \frac{3.0(52)mol}{1}*\frac{32.000g}{1mol}[/tex]        Multiply moles by molar mass.

[tex]m_{O_{2}} = \frac{3.0(52)}{1}*\frac{32.000g}{1}[/tex]             The "mol" unit cancels out.

[tex]m_{O_{2}} = 97.(6)g[/tex]                         Keep one sig. fig. to round. "6" rounds up.

[tex]m_{O_{2}} = 98g[/tex]                               <= Final answer

∴ 98 grams of oxygen are required to completely react with 110 grams of aluminum.


Related Questions

whats the chemical formula for dilute hydroxide acid​

Answers

the chemical formula is OH −

Another name for a statovolcano is a _____ volcano because it is made of alternating layers of thick, sticky lava and volcanic rock and ash. A. shield B. cinder C. composite D. pyroclastic

Answers

Answer:

Bcinder

Explanation:

URGENT!!!!
An ordered list of chemical substances is shown.

Which substances in the list can be some of the reactants and products in the same combustion reaction? (4 points)

Select one:
a. 1, 2, and 3
b. 2, 3, and 6
c. 1, 4, and 5
d. 1, 2, and 5

Answers

Answer : The correct option is (b) 2, 3, and 6

Explanation :

Combustion reactions: It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.

A balanced chemical equation is defined as the equation in which total number of individual atoms on the reactant side is equal to the total number of individual atoms on product side.

The general chemical equation for the combustion is as follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

The reactants and products in the same combustion reaction are [tex]O_2[/tex] and [tex]CO_2,H_2O[/tex] respectively.

Hence, the correct option is (b) 2, 3, and 6

Name two drying agents that can be use to dry hydrogen gas​

Answers

CaCl2, Na2SO4, MgSO4 are three drying agents that can be used to dry hydrogen gas

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two subatomic particles. Do you agree with her? Why or why not?

Answers

Answer:

The smallest constituent of an element is termed as an atom. According to Jaiden, an atom comprises two subatomic particles and three prime parts, however, this is incorrect as an atom comprises three subatomic particles and two main parts. The three subatomic particles are the protons, electrons, and neutrons. All these parts are smaller in comparison to an atom, thus, they are considered as subatomic particles.  

To be considered as an atom, an atom is needed only electrons and neutrons, like hydrogen exhibit one electron and one proton. On the other hand, the overall charge of the atom is not affected by the presence of neutrons, it is the subatomic particle that only enhances the mass of the atom.  

Answer:

I disagree with her because an atom has two main parts: the nucleus and the electron cloud. Atoms have three subatomic particles: protons, neutrons, and electrons.

Explanation:

Another chemistry question. Help me!
To prepare 250cm³ of standard solution of 1.0 mol dm-³ sodim carbonate Na2Co3​

Answers

Answer:

Molar mass Na2CO3 = 2*23 + 12 + 3*16 = 106 g/ mol

Mass of 1 mol Na2CO3 = 106 g

mass (Na2CO3) = 106g*250cm³/1000cm³ = 26.5 g

1) It is necessary because all Na2CO3 will be in the solution, and no Na2CO3 would be lost.

2) Because the solution that was formed after rinsing beaker contains Na2CO3.

3) Eyes should be on the level of calibration mark of the volumetric flask.

4) It is necessary to prevent evaporation of water, and changing , because of it, the concentration of the solution.

2 Points
A student blows a whistle. In which direction do the air particles vibrate with
respect to the direction the sound wave travels?

A. Perpendicular to

B. Parallel to

C. At a 45 degree angel to

D. At an unpredictable angel

Answers

Answer:

parallel to

Explanation:

In parallel direction the air particles vibrate with respect to the direction the sound wave travels. The correct option is B.

What is sound wave?

A sound wave is the pattern of disturbance caused by the movement of energy as it propagates away from the source of the sound through a medium (such as air, water, or any other liquid or solid matter).

Sound waves are produced by object vibrations and pressure waves, such as a ringing cellphone.

Because sound waves are produced by compressions and rarefactions in the air, they are referred to as longitudinal waves. The air particles vibrate in a direction parallel to the propagation direction.

The particles in a longitudinal wave are displaced parallel to the direction of the wave. Compressions moving along a slinky are an example of longitudinal waves.

Air particles vibrate in a parallel direction to the direction of travel of the sound wave.

Thus, the correct option is B.

For more details regarding sound waves, visit:

https://brainly.com/question/21995826

#SPJ2

PLS HELP ! 5.00 mL of a stock solution, containing 0.200 M of Na3PO4 was pipetted into a 25.00 mL volumetric flask and made up to the calibration mark with deionised water. A serial dilution was carried out for another two times to obtain the final solution of desired concentration. Calculate the concentration of the final solution, express in ppm (by volume).

Answers

Answer:

262 ppm of Na₃PO₄

Explanation:

In a dilution, the concentration of the initial solution is decreased. When you take 5.00mL of the solution that is diluted to 25.0mL The solution is diluted 25/5 = 5 times

If you make another two serial dilutions the final solution wil decrease its concentration 5*5*5 = 125 times

As original solution containing 0.200 M of Na3PO4, the final solution will have a concentration of:

0.200M / 125 = 1.6x10⁻³M

Molarity is defined as the ratio between moles and liters. 1.6x10⁻³ moles of Na3PO4 in 1L are:

1.6x10⁻³mol ₓ (164g/mol) = 0.262g Na₃PO₄ / L

Assuming density of Na3PO4 as 1g/mL the concentration of the solution is:

0.262mL Na₃PO₄ / L

As 1mL = 1000μL:

262μL Na₃PO₄ / L

μL of solute per L of solution is equal to ppm, that means the solution has:

262 ppm of Na₃PO₄

so ...this is the question and I hope you get it right one​

Answers

Answer:

About 96 grams per mole

Explanation:

The chemical formula of ammonium carbonate is [tex](NH_4)_2CO_3[/tex]. This means that breakdown of moles of atom per mole of molecule is:

2 moles of nitrogen, with a molar mass of around 14

8 moles of hydrogen, with a molar mass of around 1

1 mole of carbon, with a molar mass of about 12

3 moles of oxygen, with a molar mass of about 16

The molar mass of ammonium carbonate is therefore:

2(14)+8(1)+1(12)+3(16)=28+8+12+48=about 96

Hope this helps!

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