Question

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produced from the reaction of 6.58 g of methane and 14.4 g of oxygen gas, calculate the percent yield of water.

Answer 1

Answer: The percent yield of the water is 31.98 %

Step-by-step 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 = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

`\text{Moles of methane}=(6.58g)/(16g/mol)=0.411mol`

• For oxygen gas:

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(14.4g)/(32g/mol)=0.45mol`

The chemical equation for the combustion of methane 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, 0.45 moles of oxygen gas will react with =
`(1)/(2)* 0.45=0.225mol` 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, 0.45 moles of oxygen gas will produce =
`(2)/(2)* 0.45=0.45` moles of water

• Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

`0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol* 18g/mol)=8.1g`

• To calculate the percentage yield of water, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

`\%\text{ yield of water}=(2.59g)/(8.1g)* 100\\\\\% \text{yield of water}=31.98\%`

Hence, the percent yield of the water is 31.98 %