n ( H 2 ) = V ( H 2 ) V M = 0 ,05 L 24 L mol = 2 ,08 codt 10 − 3 mol 2 H 2 + O 2 → 2 H 2 O nitalic{ n( H_2) = { V(H_2) } over{ V_M } = { 0,05L } over { 24 L over { mol } } =2,08 codt 10^-3 mol } newline nitalic{ 2H_2 + O_2 toward 2H_2 O}