Giải thích các bước giải:
\(\begin{array}{l}
a)\\
NaBr + AgN{O_3} \to AgBr + NaN{O_3}\\
CaC{l_2} + 2AgN{O_3} \to 2AgCl + Ca{(N{O_3})_2}\\
AgN{O_3} + HCl \to AgCl + HN{O_3}\\
b)\\
{n_{AgN{O_3}}} = 0,09mol\\
{n_{AgCl}} = 0,04mol\\
\to {n_{AgN{O_3}(phảnứng)}} = 0,09 - 0,04 = 0,05mol\\
\to {n_{AgN{O_3}dư}}
\end{array}\)
Gọi:
\(\begin{array}{l}
a:C\% NaBr\\
13 - a:C\% CaC{l_2}\\
\to {m_{NaBr}} = 0,25a(g)\\
\to {m_{CaC{l_2}}} = 0,25 \times (13 - a) = 3,25 - 0,25a(g)\\
\to {n_{NaBr}} = \dfrac{{0,25a}}{{103}}mol\\
\to {n_{CaC{l_2}}} = \dfrac{{3,25 - 0,25a}}{{111}}mol\\
\to {n_{AgN{O_3}(phảnứng)}} = {n_{NaBr}} + 2{n_{CaC{l_2}}}\\
\to 0,05 = \dfrac{{0,25a}}{{103}} + \dfrac{{6,5 - 0,5a}}{{111}}\\
\to a = 4,12\\
\to {m_{NaBr}} = 1,03g\\
\to {m_{CaC{l_2}}} = 2,22g\\
\to \% {m_{NaBr}} = \dfrac{{1,03}}{{1,03 + 2,22}} \times 100\% = 31,7\% \\
\to \% {m_{CaC{l_2}}} = \dfrac{{2,22}}{{1,03 + 2,22}} \times 100\% = 68,3\%
\end{array}\)
