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.

The percent yield of the water is 31.98 %

Explanation:

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

    .....(1)

For methane:

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

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:

The chemical equation for the combustion of methane is:

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 = 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 = 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:

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

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

Hence, the percent yield of the water is 31.98 %

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