hoa tan 1800g ba(no3)2 5% vao dung dich h2so4 16% phan ung vua du a)viet phuong trinh hoa hoc b)tinh khoi luong dd thu duoc sau phan ung c)tinh nong do phann tram dd thu duoc sau phan ung

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:

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