Question

calculate the amount of water produced by the combustion of 32 g of methane

Answer 1

Equation of combustion of methane is

CH4 + 2O2 -------> CO2 + 2H2O

16 g methane on combustion gives 36 g of water

32 g methane on combustion will give 36/16 *32 = 72 g

Answer 2

Answer : The amount of water produced by the combustion of 32 g of methane is, 72 grams

Solution : Given,

Mass of methane = 32 g

Molar mass of methane = 16 g/mole

Molar mass of water = 18 g/mole

First we have to calculate the moles of methane.

$\text{Moles of methane}=\frac{\text{Mass of methane}}{\text{Molar mass of methane}}=\frac{32g}{16g/mole}=2moles$

Now we have to calculate the moles of water.

The balanced combustion reaction of methane will be,

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

From the reaction we conclude that

As, 1 mole of methane react to give 2 moles of water

So, 2 moles of methane react to give $2\times 2=4$ moles of water

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Moles of water}\times \text{Molar mass of water}$

$\text{Mass of water}=(4moles)\times (18g/mole)=72g$

Therefore, the amount of water produced by the combustion of 32 g of methane is, 72 grams