Question

Problem Page Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.2 g of butane is mixed with 32.6 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to significant digits.

Answer 1

Answer: 0.0 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of butane

$\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles$

b) moles of oxygen

$\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles$

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

According to stoichiometry :

2 moles of butane require 13 moles of $O_2$

Thus 0.09 moles of butane will require =$\frac{13}{2}\times 0.09=0.585moles$  of $O_2$

Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.

Thus all the butane will be consumed and 0.0 grams of butane will be left.