The combustion of butane can be represented by the following equation: If 256. g

Question

The combustion of butane can be represented by the following equation:

If 256. g of C4H10 and 902. g O2 react, which reactant is used up first? (enter butane or oxygen) oxygen

1. What is the mass of carbon dioxide produced? (Answer to 3 sig figs.) grams

2. What is the mass of water produced?(Answer to 3 sig figs.) grams

Calculate the number of moles of the limiting reagent. Use the stoichiometry of the balanced equation to determine how many moles of CO2 will be produced. Then convert this to grams of CO2 by using its molar mass.

Explanation / Answer

Here is the answer

Molecular mass of C4H10=58.12 g/mol

No: of moles of C4H10=256/58.12=4.4046 mol

Molecular mass of O2=32 g/mol

No: of moles of O2=902/32=28.1875

According to balanced chemical equation

2 moles of C4H10 will react with 13 moles of O2

1 moles of C4H10 will react with 6.5 moles of O2

4.4046 moles will require 4.4046*6.5=28.6299 moles of O2

But only 28.1875 moles of O2 is available making O2 as the limiting reagent

So only28.1875 moles will take part in reaction

13 moles of O2 will produce 8 moles of CO2

1 mole will produce 0.6153 moles

Hence 28.1875 moles will produce 17.23 moles

B.13 moles of O2 will produce 10 moles of H2O

1 mole will produce 0.7692 moles

Hence 28.1875 moles will produce 21.68 moles

Molar mass of H2)=18

Mass of water =390.28 g

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