Question

Butane Gas (C4H10) is used as the fuel in disposable lighters. It burns in oxygen to form carbon dioxide gas and water vapor. What mass of carbon dioxide is produced when 4.65g of Butane is burned with 10.0 g of Oxygen? How many grams of the excess reactant remain unreacted when the reaction is over?

Answer 1

Answer: The mass of unreacted butane is 1.86 g

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For butane:

Given mass of butane = 4.65 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of butane}=\frac{4.65g}{58.12g/mol}=0.080mol$

• For oxygen gas:

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{10.0g}{32g/mol}=0.3125mol$

The chemical equation for the combustion of butane follows:

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

By Stoichiometry of the reaction:

13 moles of oxygen gas reacts with 2 moles of butane

So, 0.3125 moles of oxygen gas will react with = $\frac{2}{13}\times 0.3125=0.048mol$ of butane

As, given amount of butane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

Moles of butane remained = (0.080 - 0.048) = 0.032 moles

Now, calculating the mass of butane using equation 1, we get:

Moles of butane unreacted = 0.032 moles

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

$0.032mol=\frac{\text{Mass of butane}}{58.12g/mol}\\\\\text{Mass of butane}=(0.032mol\times 58.12g/mol)=1.86g$

Hence, the mass of unreacted butane is 1.86 g