Question

8.50g of CO when heated in a current of dry hydrogen gave 6.85g of CU and 2.16g f H2O. calculate the proportion of hydrogen to oxygen by mass in water.

Answer 1

The equation for the reaction is;

$CuO_{(s)}+H_{2{(g)}}\to Cu_{(s)}+H_2O_{(g)}$

Mass of $Cu_{(g)}$ which reacts from $8.50g$ of $CuO_{(s)}$ $=6.85g$

Mass of $O_{2{(g)}}$ from the reaction $=$ $8.50g-6.85g=1.65g$

From the law of constant proportions, which states that in a chemical substance, the elements are always present in definite proportions by mass, the mass of $H_{2(g)}$ in water formed from the reaction is;

$2.16g H_2O-1.65O_2=0.51gH_2$

The proportion of $H_2$ to $O_2$ $=0.51 g$ to $1.65g$

Representing a ratio of $H_2:O_2=1:3$