Giải thích các bước giải:

\(\begin{array}{l}
A.\\
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}\\
{n_{Ba}} = 0,25mol\\
 \to {n_{Ba{{(OH)}_2}}} = {n_{{H_2}}} = {n_{Ba}} = 0,25mol\\
 \to {m_{Ba{{(OH)}_2}}} = 0,25 \times 171 = 42,75g\\
 \to {m_{{\rm{dd}}Ba{{(OH)}_2}}} = {m_{Ba}} + {m_{{H_2}O}} - {m_{{H_2}}} = 34,25 + 66,35 - 0,25 \times 2\\
 \to {m_{{\rm{dd}}Ba{{(OH)}_2}}} = 100,1g\\
 \to C{\% _{{\rm{dd}}Ba{{(OH)}_2}}} = \dfrac{{42,75}}{{100,1}} \times 100\%  = 42,7\% 
\end{array}\)

B,

\(\begin{array}{l}
Ba + 2{H_2}O \to Ba{(OH)_2} + {H_2}\\
{n_{Ba}} = {n_{Ba{{(OH)}_2}}} = 0,01mol\\
 \to {m_{Ba{{(OH)}_2}}} = 1,71g\\
 \to {m_{Ba{{(OH)}_2}}} = \dfrac{{200 \times 8,55}}{{100}} = 17,1g\\
 \to C\% {m_{Ba{{(OH)}_2}}} = \dfrac{{17,1 + 1,71}}{{200}} \times 100\%  = 9,405\% 
\end{array}\)

Đáp án:

Giải thích các bước giải:

A.Ba+2H2O→Ba(OH)2+H2nBa=0,25mol→nBa(OH)2=nH2=nBa=0,25mol→mBa(OH)2=0,25×171=42,75g→mddBa(OH)2=mBa+mH2O−mH2=34,25+66,35−0,25×2→mddBa(OH)2=100,1g→C%ddBa(OH)2=42,75100,1×100%=42,7%A.Ba+2H2O→Ba(OH)2+H2nBa=0,25mol→nBa(OH)2=nH2=nBa=0,25mol→mBa(OH)2=0,25×171=42,75g→mddBa(OH)2=mBa+mH2O−mH2=34,25+66,35−0,25×2→mddBa(OH)2=100,1g→C%ddBa(OH)2=42,75100,1×100%=42,7%

B,

Ba+2H2O→Ba(OH)2+H2nBa=nBa(OH)2=0,01mol→mBa(OH)2=1,71g→mBa(OH)2=200×8,55100=17,1g→C%mBa(OH)2=17,1+1,71200×100%=9,405%Ba+2H2O→Ba(OH)2+H2nBa=nBa(OH)2=0,01mol→mBa(OH)2=1,71g→mBa(OH)2=200×8,55100=17,1g→C%mBa(OH)2=17,1+1,71200×100%=9,405%

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