Giải thích các bước giải:
\(\begin{array}{l}
34.\\
F{e_2}{O_3} + 3CO \to 2Fe + 3C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{m_{F{e_2}{O_3}(tt)}} = \dfrac{{24 \times 100\% }}{{80\% }} = 30g\\
{n_{F{e_2}{O_3}}} = 0,19mol\\
\to {n_{C{O_2}}} = 3{n_{F{e_2}{O_3}}} = 0,57mol\\
\to {n_{Fe}} = 2{n_{F{e_2}{O_3}}} = 0,38mol\\
\to {n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,57mol \to {m_{CaC{O_3}}} = 57g\\
{n_{{H_2}}} = {n_{Fe}} = 0,38mol \to {V_{{H_2}}} = 8,512l\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,38mol \to {m_{FeC{l_2}}} = 48,26g\\
\to {m_{{\rm{dd}}B1}} = {m_{Fe}} + {m_{HCl}} - {m_{{H_2}}} = 520,52g\\
\to C{\% _{FeC{l_2}}} = \dfrac{{48,26}}{{520,52}} \times 100\% = 9,27\% \\
35.\\
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
Ba{(OH)_2} + 2HCl \to BaC{l_2} + 2{H_2}O\\
b)\\
{n_{HCl}} = 0,5mol\\
{n_{Ba{{(OH)}_2}}} = 0,1mol\\
\to {n_{HCl}}(2) = 2{n_{Ba{{(OH)}_2}}} = 0,4mol\\
\to {n_{HCl}}(1) = 0,1mol\\
\to {n_{N{a_2}C{O_3}}} = \dfrac{1}{2}{n_{HCl}}(1) = 0,05mol\\
\to a = {m_{N{a_2}C{O_3}}} = 5,3g\\
\to {n_{C{O_2}}} = \dfrac{1}{2}{n_{HCl}}(1) = 0,05mol \to {V_{C{O_2}}} = 1,12l
\end{array}\)