Đáp án:
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Giải thích các bước giải:
a,
\(\begin{array}{l}
NaCl + AgN{O_3} \to AgCl + NaN{O_3}\\
{n_{NaCl}} = 0,5mol\\
{n_{AgN{O_3}}} = 0,3mol\\
{n_{NaCl}} > {n_{AgN{O_3}}}
\end{array}\)
Suy ra NaCl dư
\(\begin{array}{l}
\to {n_{NaN{O_3}}} = {n_{AgN{O_3}}} = 0,3mol\\
\to {n_{NaCl}}dư= 0,5 - 0,3 = 0,2mol\\
\to {n_{N{a^ + }}} = {n_{NaN{O_3}}} + {n_{NaCl}}dư= 0,5mol\\
\to {n_{N{O_3}^ - }} = {n_{AgN{O_3}}} = 0,3mol\\
\to {n_{C{l^ - }}} = {n_{NaCl}}dư= 0,2mol\\
\to C{M_{NaN{O_3}}} = \dfrac{{0,3}}{{0,5 + 0,3}} = 0,375M\\
\to C{M_{NaCl}}dư= \dfrac{{0,2}}{{0,5 + 0,3}} = 0,25M\\
\to C{M_{N{a^ + }}} = \dfrac{{0,5}}{{0,5 + 0,3}} = 0,625M\\
\to C{M_{N{O_3}^ - }} = C{M_{NaN{O_3}}} = \dfrac{{0,3}}{{0,5 + 0,3}} = 0,375M\\
\to C{M_{C{l^ - }}} = C{M_{NaCl}}dư= \dfrac{{0,2}}{{0,5 + 0,3}} = 0,25M
\end{array}\)
b,
\(\begin{array}{l}
N{a_2}C{O_3} + BaC{l_2} \to 2NaCl + BaC{O_3}\\
{n_{N{a_2}C{O_3}}} = 0,2mol\\
{n_{BaC{l_2}}} = 0,3mol\\
\to {n_{BaC{l_2}}} > {n_{N{a_2}C{O_3}}}
\end{array}\)
Suy ra \(BaC{l_2}\) dư
\(\begin{array}{l}
\to {n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,4mol\\
\to {n_{BaC{l_2}}}dư= 0,3 - 0,2 = 0,1mol\\
\to {n_{C{l^ - }}} = {n_{NaCl}} + 2{n_{BaC{l_2}}}dư= 0,6mol\\
\to C{M_{NaCl}} = \dfrac{{0,4}}{{0,2 + 0,3}} = 0,8M = C{M_{N{a^ + }}}\\
\to C{M_{BaC{l_2}}}dư= \dfrac{{0,1}}{{0,2 + 0,3}} = 0,2M = C{M_{B{a^{2 + }}}}\\
\to C{M_{C{l^ - }}} = \dfrac{{0,6}}{{0,2 + 0,3}} = 1,2M
\end{array}\)
c,
\(\begin{array}{l}
Ca{(OH)_2} + 2HCl \to CaC{l_2} + 2{H_2}O\\
{n_{HCl}} = 0,5mol\\
{n_{Ca{{(OH)}_2}}} = 0,35mol\\
\dfrac{{{n_{HCl}}}}{2} < {n_{Ca{{(OH)}_2}}}
\end{array}\)
Suy ra \(Ca{(OH)_2}\)
\(\begin{array}{l}
\to {n_{CaC{l_2}}} = \dfrac{1}{2}{n_{HCl}} = 0,25mol\\
\to {n_{Ca{{(OH)}_2}}}dư= 0,35 - \dfrac{1}{2} \times 0,5 = 0,1mol\\
\to {n_{C{a^{2 + }}}} = {n_{CaC{l_2}}} + {n_{Ca{{(OH)}_2}}}dư= 0,35mol\\
\to {n_{C{l^ - }}} = 2{n_{CaC{l_2}}} = 0,5mol\\
\to {n_{O{H^ - }}} = 2{n_{Ca{{(OH)}_2}}}dư= 0,2mol\\
\to C{M_{C{a^{2 + }}}} = \dfrac{{0,35}}{{0,25 + 0,35}} = 0,583M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,5}}{{0,25 + 0,35}} = 0,833M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,2}}{{0,25 + 0,35}} = 0,333M
\end{array}\)
