Đáp án:
\(\begin{array}{l}
b)\\
{m_{{\rm{dd}}spu}} = 1930,9275g\\
c)\\
{C_\% }HN{O_3} = 2,25\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Ba{(N{O_3})_2} + {H_2}S{O_4} \to 2HN{O_3} + BaS{O_4}\\
b)\\
{n_{Ba{{(N{O_3})}_2}}} = \dfrac{{1800 \times 5\% }}{{261}} = 0,345\,mol\\
{n_{{H_2}S{O_4}}} = {n_{BaS{O_4}}} = {n_{Ba{{(N{O_3})}_2}}} = 0,345\,mol\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{0,345 \times 98}}{{16\% }} = 211,3125g\\
{m_{{\rm{dd}}spu}} = 1800 + 211,3125 - 0,345 \times 233 = 1930,9275g\\
c)\\
{n_{HN{O_3}}} = 2{n_{Ba{{(N{O_3})}_2}}} = 0,69\,mol\\
{C_\% }HN{O_3} = \dfrac{{0,69 \times 63}}{{1930,9275}} \times 100\% = 2,25\%
\end{array}\)
