Gọi `x` là số mol `Ba`

`n_(H_2)=\frac{8,96}{22,4}=0,4(mol)`

`n_(Al (dư))=\frac{5,4}{27}=0,2(mol)`

`Ba+2H_2O->Ba(OH)_2+H_2`

`Ba(OH)_2+2H_2O+2Al->Ba(AlO_2)_2+3H_2`

Theo PT

`n_(H_2)=x+3x=0,4`

`=>x=0,1(mol)`

Theo PT

`n_(Al  (pứ))=2x=0,2(mol)`

$⇒∑n_(Al)=0,2+0,2=0,4(mol)$

`=>m_(Al)=0,4.27=10,8(g)`

`m_(Ba)=0,1.137=13,7(g)`

`=>m=m_(Ba)+m_(Al)=10,8+13,7=24,5(g)`

`%m_(Ba)=\frac{13,7}{24,5}.100=55,92%`

`%m_(Al)=100-55,92=44,08%`

Theo PT

`n_(Ba(AlO_2)_2)=n_(Ba)=0,1(mol)`

`m_(dd)=19,1+200-0,4.2=218,3(g)`

`=>C%_(Ba(AlO_2)_2)=\frac{0,1.255}{218,3}.100=11,68%`

Đáp án:

\( \% {m_{Ba}} = 55,9\% ; \% {m_{Al}} = 44,1\% \)

\( C{\% _{Ba{{(Al{O_2})}_2}}}= 11,68\% \)

Giải thích các bước giải:

 Rắn không tan \(Al\) dư.

Vậy phản ứng xảy ra:

\(Ba + 2{H_2}O\xrightarrow{{}}Ba{(OH)_2} + {H_2}\)

\(Ba{(OH)_2} + 2Al + 2{H_2}O\xrightarrow{{}}Ba{(Al{O_2})_2} + 3{H_2}\)

Gọi số mol \(Ba\) là \(x\)

\( \to {n_{Ba{{(OH)}_2}}} = {n_{Ba}} = x\)

\( \to {n_{{H_2}}} = {n_{Ba{{(OH)}_2}}} + 3{n_{Ba{{(OH)}_2}}} = 4x = \frac{{8,96}}{{22,4}} = 0,4{\text{ mol}}\)

\( \to x = 0,1\)

Ta có:

\({n_{Ba}} = 0,1{\text{ mol;}}{{\text{n}}_{Al{\text{ phản ứng}}}} = 2{n_{Ba{{(OH)}_2}}} = 0,2{\text{ mol}}\)

\( \to {m_{Ba}} = 0,1.137 = 13,7{\text{ gam}}\)

\({m_{Al}} = 0,2.27 + 5,4 = 10,8{\text{ gam}}\)

\( \to \% {m_{Ba}} = \frac{{13,7}}{{13,7 + 10,8}} = 55,9\%  \to \% {m_{Al}} = 44,1\% \)

BTKL:

\({m_{Ba}} + {m_{Al}} + {m_{{H_2}O}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 0,1.137 + 0,2.27 + 200 = {m_{dd}} + 0,4.2\)

\( \to {m_{dd}} = 218,3{\text{ gam}}\)

\({n_{Ba{{(Al{O_2})}_2}}} = {n_{Ba}} = 0,1{\text{ mol}}\)

\( \to {m_{Ba{{(Al{O_2})}_2}}} = 0,1.(137 + 27.2 + 16.4) = 25,5{\text{ gam}}\)

\( \to C{\% _{Ba{{(Al{O_2})}_2}}} = \frac{{25,5}}{{218,3}} = 11,68\% \)