Question

Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) . If 0.122g of water is produced from the reaction of 0.16g of methane and 0.25g of oxygen gas, calculate the percent yield of water.

Answer 1

Answer: The percent yield of water is 84.72 %

Explanation:

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

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

For methane:

Given mass of methane= 0.16 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{0.16g}{16g/mol}=0.01mol$

For oxygen gas:

Given mass of oxygen gas = 0.25 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{0.25g}{32g/mol}=0.008mol$

The chemical equation for the reaction is

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.008 moles of oxygen gas will react with = $\frac{1}{2}\times 0.008=0.004mol$ of methane

As, given amount of methane 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.

By Stoichiometry of the reaction:

2 moles of oxygen gas produces = 2 moles of water

So, 0.008 moles of oxygen gas will produce = $\frac{2}{2}\times 0.008=0.008moles$ of water

Mass of water=$moles\times {\text{Molar Mass}}=0.008\times 18=0.144g$

To calculate the percentage yield of titanium (IV) chloride, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 0.122

Theoretical yield of water = 0.144 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{0.122g}{0.144g}\times 100\\\\\% \text{yield of water}=84.72\%$

Hence, the percent yield of water is 84.72%.