Đáp án:
\(\begin{array}{l}
1)\\
b)\\
{m_{KCl}} = 11,175g\\
c)\\
{C_\% }KOH = 2,62\% \\
{C_\% }KCl = 5,22\% \\
2)\\
b)\\
{m_{CaC{O_3}}} = 20g\\
c)\\
{C_\% }{K_2}C{O_3} = 7,81\% \\
{C_\% }KCl = 5,62\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\\
a)\\
KOH + HCl \to KCl + {H_2}O\\
b)\\
{n_{KOH}} = \dfrac{{14}}{{56}} = 0,25\,mol\\
{n_{HCl}} = \dfrac{{200 \times 2,7375\% }}{{36,5}} = 0,15\,mol\\
\dfrac{{0,15}}{1} < \dfrac{{0,25}}{1} \Rightarrow \text{ KOH dư}\\
{n_{KCl}} = {n_{HCl}} = 0,15\,mol\\
{m_{KCl}} = 0,15 \times 74,5 = 11,175g\\
c)\\
{m_{{\rm{dd}}spu}} = 14 + 200 = 214g\\
{n_{KOH}} \text{ dư}= 0,25 - 0,15 = 0,1\,mol\\
{C_\% }KOH = \dfrac{{0,1 \times 56}}{{214}} \times 100\% = 2,62\% \\
{C_\% }KCl = \dfrac{{11,175}}{{214}} \times 100\% = 5,22\% \\
2)\\
a)\\
CaC{l_2} + {K_2}C{O_3} \to 2KCl + CaC{O_3}\\
b)\\
{n_{CaC{l_2}}} = \dfrac{{250 \times 8,88\% }}{{111}} = 0,2\,mol\\
{n_{{K_2}C{O_3}}} = \dfrac{{300 \times 23\% }}{{138}} = 0,5\,mol\\
\dfrac{{0,2}}{1} < \dfrac{{0,5}}{1} \Rightarrow \text{ $K_2CO_3$ dư}\\
{n_{CaC{O_3}}} = {n_{CaC{l_2}}} = 0,2\,mol\\
{m_{CaC{O_3}}} = 0,2 \times 100 = 20g\\
c)\\
{m_{{\rm{dd}}spu}} = 250 + 300 - 20 = 530g\\
{C_\% }{K_2}C{O_3} = \dfrac{{(0,5 - 0,2) \times 138}}{{530}} \times 100\% = 7,81\% \\
{n_{KCl}} = 2{n_{CaC{l_2}}} = 0,4\,mol\\
{C_\% }KCl = \dfrac{{0,4 \times 74,5}}{{530}} \times 100\% = 5,62\%
\end{array}\)
