\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Mg + 2CH3COOH ---> (CH3COO)2Mg + H2
0,15<-------------------------------------------0,15
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,15.24=3,6\left(g\right)\\m_{MgO}=7,6-3,6=4\left(g\right)\end{matrix}\right.\\ \Rightarrow B\)
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