Question

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

Answer 1

CH4 + 2 O2 = CO2 + 2 H2O
1 mol 2mol 1mol 2mol ( by avagardo hypothesis )

now 16g of methane is 1 mol of methane

thus amount of water in 2 mol = 2 * 18 = 36g

so 16g of methane produces 36g of water.

Answer 2

Answer: 72 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

$\text{Number of moles of methane}=\frac{32g}{16}=2moles$  

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

Accordig to stoichiometry:

1 mole of $CH_4$ gives = 2 moles of water

Thus 2 moles of $CH_4$ gives=$\frac{2}{1}\times 2=4$ moles of water

Mass of water=$moles\times Molar Mass=4\times 18=72g$