Then, $H_2C_2O_4 = (10-x)\,gm$

$\therefore \,{n_{NaHC{O_3}}} = \frac{x}{{84}}$

$2NaHC{O_3} \to N{a_2}C{O_3} + {H_2}O + C{O_2}$

$\therefore {n_{C{O_2}}} = \frac{x}{{168}}$

Total $C{O_2} = \frac{x}{{168}} + \frac{{10 - x}}{{90}} = \frac{{0.25}}{{25}}$

On solving $'x'$

              $\%  = \frac{x}{{10}} \times 100 = 10\,x$

${n_{{H_2}{C_2}{O_4}}} = \left( {\frac{{10 - x}}{{90}}} \right)$

${H_2}{C_2}{O_4} \to {H_2}O + C{O_2} + CO$

$\therefore {n_{C{O_2}}} = \left( {\frac{{10 - x}}{{90}}} \right)$