Giải thích các bước giải:

 Bài 3:

\(\begin{array}{l}
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
a)\\
{n_{N{a_2}C{O_3}}} = 0,24mol\\
 \to {n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,48mol\\
 \to {m_{HCl}} = 0,48 \times 36,5 = 17,52g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{17,52}}{{15\% }} \times 100\%  = 116,8g\\
b)\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,24mol\\
 \to {V_{C{O_2}}} = 0,24 \times 22,4 = 5,376l\\
c)\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,48mol\\
 \to {m_{NaCl}} = 0,48 \times 58,5 = 28,08g\\
 \to {m_{{\rm{dd}}NaCl}} = {m_{N{a_2}C{O_3}}} + {m_{{\rm{dd}}HCl}} - {m_{C{O_2}}} = 25 + 116,8 - 0,24 \times 60\\
 \to {m_{{\rm{dd}}NaCl}} = 127,4g\\
 \to C{\% _{{\rm{dd}}NaCl}} = \dfrac{{28,08}}{{127,4}} \times 100\%  = 22,04\% 
\end{array}\)

Bài 4:

\(\begin{array}{l}
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
a)\\
{n_{Al}} = 0,2mol\\
 \to {n_{{H_2}S{O_4}}} = \dfrac{3}{2}{n_{Al}} = 0,3mol\\
 \to {m_{{H_2}S{O_4}}} = 0,3 \times 98 = 29,4g\\
 \to {m_{{\rm{dd}}}}_{{H_2}S{O_4}} = \dfrac{{29,4}}{{20\% }} \times 100\%  = 147g\\
b)\\
{n_{{H_2}}} = \frac{3}{2}{n_{Al}} = 0,3mol\\
 \to {m_{{H_2}}} = 0,6g\\
 \to {n_{A{l_2}{{(S{O_4})}_3}}} = \dfrac{1}{2}{n_{Al}} = 0,1mol\\
 \to {m_{A{l_2}{{(S{O_4})}_3}}} = 0,1 \times 342 = 34,2g\\
 \to {m_{{\rm{dd}}A{l_2}{{(S{O_4})}_3}}} = {m_{Al}} + {m_{{\rm{dd}}{H_2}S{O_4}}} - {m_{{H_2}}} = 151,8g\\
 \to C{\% _{A{l_2}{{(S{O_4})}_3}}} = \dfrac{{34,2}}{{151,8}} \times 100\%  = 22,53\% 
\end{array}\)