Question

How many grams of water will be produced by complete combustion of 12g of methane gas​

Answer 1

Mass of water produced by complete combustion of 12 gram of methane gas is 27 grams.

Answer 2

Answer: The mass of water produced is 27 grams.

Explanation:

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

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

Given mass of methane gas = 12 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{12g}{16g/mol}=0.75mol$

The chemical equation for the combustion of methane follows:

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

By Stoichiometry of the reaction:

1 mole of methane gas produces 2 moles of water.

So, 0.75 moles of methane gas will produce = $\frac{2}{1}\times 0.75=1.5mol$ of water.

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

Molar mass of water = 18 g/mol

Moles of water = 1.5 moles

Putting values in equation 1, we get:

$1.5mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.5mol\times 18g/mol)=27g$

Hence, the mass of water produced is 27 grams.