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Q.
The amount of water $(g)$ produced by the combustion of 16 g of methane is
Some Basic Concepts of Chemistry
Solution:
The balanced equation for combustion of methane is $CH_{4\left(g\right)}+2O_{2\left(g\right)}+2H_{2}O_{\left(g\right)}$
(i) $16\, g$ of $CH_4$ corresponds to one mole.
(ii) From the above equation, $1 \,mol$ of $CH_{4(g)}$ gives $2\, mol$ of $H_2O_{(g)}.$
$2 \,mol$ of water $(H_2O) = 2 × (2 + 16) = 2 × 18 = 36\, g$
$1\, mol\, H_{2}O = 18 \,g\, H_{2}O \Rightarrow \frac{18\,g\,H_{2}O}{1\,mol\,H_{2}O}=1$
Hence $2\, mol \,H_{2}O\times\frac{18\,g\,H_{2}O}{1\,mol\,H_{2}O}=2\times8\,H_{2}O=36\,g\,H_{2}O.$