m gam hỗn hợp A\(\left\{{}\begin{matrix}Na:a\left(mol\right)\\Ca:2a\left(mol\right)\end{matrix}\right.\)
\(2Na\left(a\right)+2CH_3COOH--->2CH_3COONa\left(a\right)+H_2\left(0,5a\right)\)\(\left(1\right)\)
\(Ca\left(2a\right)+2CH_3COOH--->\left(CH_3COO\right)_2Ca\left(2a\right)+H_2\left(2a\right)\)\(\left(2\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PTHH (1) và (2) \(\sum n_{H_2}=0,5a+2a=2,5a\left(mol\right)\)
\(\Rightarrow2,5a=0,4\)
\(\Rightarrow a=0,16\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,16.23=3,68\left(g\right)\\m_{Ca}=2.0,16.40=12,8\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m=16,48\left(g\right)\)
Đến đây thì đơn giản, sẽ tính được phần trăm khối lượng mỗi chất
\(m_{H_2}=0,4.2=0,8\left(g\right)\)
\(m_{ddCH_3COOH}=240.1,25=300\left(g\right)\)
\(m_{ddB}=16,48+300-0,8=315,68\left(g\right)\)
Theo PTHH (1) và (2) \(\left\{{}\begin{matrix}n_{CH_3COONa}=a=0,16\left(mol\right)\\n_{\left(CH_3COO\right)_2Ca}=2a=0,32\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,16.82}{315,68}.100\%=4,16\%\)
\(C\%_{\left(CH_3COO\right)_2Ca}=\dfrac{0,32.158}{315,68}.100\%=16,02\%\)
