a, \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
b, Ta có: 24nMg + 65nZn = 4,245 (1)
Theo PT: \(n_{CH_3COOH}=2n_{Mg}+2n_{Zn}=0,2.1=0,2\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,055\left(mol\right)\\n_{Zn}=0,045\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,055.24=1,32\left(g\right)\\m_{Zn}=0,045.65=2,925\left(g\right)\end{matrix}\right.\)
c, \(n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
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