a
Let \(NaHCO_3 = x\,gm\)

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 \('x'\)

              \(\%  = \frac{x}{{10}} \times 100 = 10\,x\)

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

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

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