\(n_{Zn}=0,1\left(mol\right)\)
\(n_{HCl}=0,4\left(mol\right)\)
\(Zn+2HCl-->ZnCl_2+H_2\uparrow\)
\(\dfrac{0,1}{1}< \dfrac{0,4}{2}\) => Zn hết HCl dư
a) \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
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