Đáp án:

\(\begin{array}{l}
C{\% _{{K_2}S{O_4}}} = 3,484\% \\
C{\% _{{H_2}S{O_4}}} = 7,85\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{K_2}C{O_3} + {H_2}S{O_4} \to {K_2}S{O_4} + C{O_2} + {H_2}O\\
{n_{{K_2}C{O_3}}} = \dfrac{{13,8}}{{138}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{490 \times 10}}{{100}} = 49g\\
{n_{{H_2}S{O_4}}} = \dfrac{{4,9}}{{98}} = 0,5mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,5}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{C{O_2}}} = {n_{{K_2}C{O_3}}} = 0,1mol\\
{m_{C{O_2}}} = 0,1 \times 44 = 4,4g\\
{m_{{\rm{dd}}spu}} = 13,8 + 490 - 4,4 = 499,4g\\
{n_{{K_2}S{O_4}}} = {n_{{K_2}C{O_3}}} = 0,1mol\\
{m_{{K_2}S{O_4}}} = 0,1 \times 174 = 17,4g\\
{n_{{H_2}S{O_4}d}} = 0,5 - 0,1 = 0,4mol\\
{m_{{H_2}S{O_4}d}} = 0,4 \times 98 = 39,2g\\
C{\% _{{K_2}S{O_4}}} = \dfrac{{17,4}}{{499,4}} \times 100\%  = 3,484\% \\
C{\% _{{H_2}S{O_4}}} = \dfrac{{39,2}}{{499,4}} \times 100\%  = 7,85\% 
\end{array}\)