\(n_{Zn}=\dfrac{32.5}{65}=0.5\left(mol\right)\)
\(n_{HCl}=\dfrac{29.2}{36.5}=0.8\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(1.........2\)
\(0.5......0.8\)
\(LTL:\dfrac{0.5}{1}>\dfrac{0.8}{2}\Rightarrow Zndư\)
\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)
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