\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Ba + 2H2O --> Ba(OH)2 + H2
0,1<------------------------0,1
=> mBa = 0,1.137 = 13,7 (g)
=> mCu = 20 - 13,7 = 6,3 (g)
\(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{13,7}{20}.100\%=68,5\%\\\%m_{Cu}=\dfrac{6,3}{20}.100\%=31,5\%\end{matrix}\right.\)
Ba+2H2O->Ba(OH)2+H2
0,1--------------------------0,1 mol
n H2 =\(\dfrac{2,24}{22,4}=0,1mol\)
m Ba=0,1.137=13,7g=>%Ba=68,5%
=>m Cu=20-13,7=6,3g=>%Cu=31,5%
Vik chỉ có Ba tác dụng đc vs nước (H2O) , Cu ko td vs nước nên ta có :
PTHH : \(Ba+2H_2O->Ba\left(OH\right)_2+H_2\uparrow\) (1)
\(n_{H_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Từ (1) -> \(n_{Ba}=n_{H_2}=0,1\left(mol\right)\)
-> \(m_{Ba}=n.M=0,1.137=13,7\left(g\right)\)
=> \(\%m_{Ba}=\dfrac{m_{Ba}}{m_{HH}}.100\%=68,5\%\)
\(\%m_{Cu}=100\%-\%m_{Ba}=100\%-68,5\%=31,5\%\)
