Đáp án:
1,\({m_{BaS{O_4}}} = 34,95g\)
2,\({m_{BaS{O_4}}} = 46,6g\)
Giải thích các bước giải:
1,
\(\begin{array}{l}
3Ba{(OH)_2} + A{l_2}{(S{O_4})_3} \to 3BaS{O_4} + 2Al{(OH)_3}\\
{n_{Ba{{(OH)}_2}}} = 0,25mol\\
{n_{A{l_2}{{(S{O_4})}_3}}} = 0,05mol\\
\dfrac{{{n_{Ba{{(OH)}_2}}}}}{3} > {n_{A{l_2}{{(S{O_4})}_3}}}
\end{array}\)
Suy ra \(Ba{(OH)_2}\)
\(\begin{array}{l}
\to {n_{Ba{{(OH)}_2}}}dư= 0,25 - 3 \times 0,05 = 0,1mol\\
\to {n_{Al{{(OH)}_3}}} = 2 \times 0,05 = 0,1mol\\
2Al{(OH)_3} + Ba{(OH)_2} \to Ba{(Al{O_2})_2} + 4{H_2}O\\
\dfrac{{{n_{Al{{(OH)}_3}}}}}{2} < {n_{Ba{{(OH)}_2}}}dư
\end{array}\)
Vậy kết tủa chỉ còn có:\(BaS{O_4}\)
\(\begin{array}{l}
\to {n_{BaS{O_4}}} = 3{n_{A{l_2}{{(S{O_4})}_3}}} = 0,15mol\\
\to {m_{BaS{O_4}}} = 34,95g
\end{array}\)
2,
\(\begin{array}{l}
{n_{KAl(S{O_4}).12{H_2}O}} = 0,1mol\\
{n_{Ba{{(OH)}_2}}} = 0,2mol
\end{array}\)
Ta có các phương trình ion sau:
\(\begin{array}{l}
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
A{l^{3 + }} + 3O{H^ - } \to Al{(OH)_3}\\
{n_{A{l^{3 + }}}} = {n_{KAl(S{O_4}).12{H_2}O}} = 0,1mol\\
{n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} = 0,4mol\\
\dfrac{{{n_{O{H^ - }}}}}{3} > {n_{A{l^{3 + }}}}
\end{array}\)
Suy ra \(O{H^ - }\) dư
\(\begin{array}{l}
Al{(OH)_3} + O{H^ - } \to Al{O_2}^ - + {H_2}O\\
{n_{Al{{(OH)}_3}}} = {n_{A{l^{3 + }}}} = 0,1mol\\
{n_{O{H^ - }}}dư= 0,4 - 3 \times 0,1 = 0,1mol\\
\to {n_{Al{{(OH)}_3}}} = {n_{O{H^ - }}}dư
\end{array}\)
vậy kết tủa chỉ còn \(BaS{O_4}\)
\(\begin{array}{l}
B{a^{2 + }} + S{O_4}^{2 - } \to BaS{O_4}\\
{n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,2mol\\
{n_{S{O_4}^{2 - }}} = 2{n_{KAl{{(S{O_4})}_2}.12{H_2}O}} = 0,2mol\\
\to {n_{BaS{O_4}}} = 0,2mol\\
\to {m_{BaS{O_4}}} = 46,6g
\end{array}\)
