The amount of water (g) produced by the combustion of 16 g of methane is

Question

The amount of water (g) produced by the combustion of 16 g of methane is (A) 16 g (B) 36 g (C) 18 g (D) 32 g

Tardigrade Admin · Accepted Answer

The balanced equation for combustion of methane is

C H_{4 (g)} + 2 O_{2 (g)} + 2 H_{2} O_{(g)}

(i)

16 g

of

C H_{4}

corresponds to one mole.
(ii) From the above equation,

1 m o l

of

C H_{4 (g)}

gives

2 m o l

of

H_{2} O_{(g)} .

2 m o l

of water

(H_{2} O) = 2 \times (2 + 16) = 2 \times 18 = 36 g

1 m o l H_{2} O = 18 g H_{2} O \Rightarrow \frac{18 g H _{2} O}{1 m o l H _{2} O} = 1

Hence

2 m o l H_{2} O \times \frac{18 g H _{2} O}{1 m o l H _{2} O} = 2 \times 8 H_{2} O = 36 g H_{2} O .

Q. The amount of water $(g)$ produced by the combustion of 16 g of methane is