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

Step-by-step 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}=(5.2g)/(58.12g/mol)=0.09moles`

b) moles of oxygen

`\text{Number of moles}=(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 =
`(13)/(2)* 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.