Chemistry High School

2C2H6 + 7O2 ------> 4CO2 + 6H2O


If you are given 10 grams of C2H6 and 10 grams of O2 how many grams of H2O would you produce? Hint: find the limiting reactant. Answer should have 3 decimal places.

Right answer gets brainliest

Answers

Answer 1

We convert the masses of our reactants to moles and use the stoichiometric coefficients to determine which one of our reactants will be limiting.

Dividing the mass of each reactant by its molar mass:

(10 g C2H6)(30.069 g/mol) = 0.3326 mol C2H6

(10 g O2)(31.999 g/mol) = 0.3125 mol O2.

Every 2 moles of C2H6 react with 7 moles of O2. So the number of moles of O2 needed to react completely with 0.3326 mol C2H6 would be (0.3326)(7/2) = 1.164 mol O2. That is far more than the number of moles of O2 that we are given: 0.3125 moles. Thus, O2 is our limiting reactant.

Since O2 is the limiting reactant, its quantity will determine how much of each product is formed. We are asked to find the number of grams (the mass) of H2O produced. The molar ratio between H2O and O2 per the balanced equation is 6:7. That is, for every 6 moles of H2O that is produced, 7 moles of O2 is used up (intuitively, then, the number of moles of H2O produced should be less than the number of moles of O2 consumed).

So, the number of moles of H2O produced would be (0.3125 mol O2)(6 mol H2O/7 mol O2) = 0.2679 mol H2O. We multiply by the molar mass of H2O to convert moles to mass: (0.2679 mol H2O)(18.0153 g/mol) = 4.826 g H2O.

Given 10 grams of C2H6 and 10 grams of O2, 4.826 g of H2O are produced.

Related Questions

a 5 g ice cube starts life at -1 C if 2182.5 j are added to it what will be the final temperature upon his demise ?

Answers

Answer:

Final Temp = 23.92°C

Explanation:

ΔH(total) = (m·c·ΔT)ice + (m·ΔH(f))melt'g + (m·c·ΔT)water

2182.5j = (5g)(2.092j/g·°C)(1°C) + (5g)(334.56j/g) + (5g)(4.184j/g·°C)(ΔT)

(2182.5 - 10.46 - 1672.8)j = 20.92j/°C·ΔT

ΔT = (2182.5 - 10.46 - 1672.8)j / 20.92j/°C = 23.92°C

Since the melting ice starts and ends at 0°C and is then warmed to 23.92°C then the temperature change is also the final temp of the water based upon given energy input values.


A reaction in the laboratory yields 5.98 g KAI(SO2)2, How many moles of potassium aluminum
sulfate were produced?

Answers

Answer:

0.0308 mol

Explanation:

In order to convert from grams of any given substance to moles, we need to use its molar mass:

Molar mass of KAI(SO₂)₂ = MM of K + MM of Al + (MM of S + 2*MM of O)*2Molar mass of KAI(SO₂)₂ = 194 g/mol

Now we calculate the number of moles of KAI(SO₂)₂ contained in 5.98 g:

5.98 g ÷ 194 g/mol = 0.0308 mol
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