Mg + H2SO4 → MgSO4 + H2
\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
a) Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)
b) Theo pT: \(n_{H_2SO_4}=0,1\times98=9,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{9,8}{30\%}=32,67\left(g\right)\)
c) \(m_{H_2}=0,1\times2=0,2\left(g\right)\)
Ta có: \(m_{dd}saupư=2,4+32,67-0,2=34,87\left(g\right)\)
d) Theo pT: \(n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgSO_4}=0,1\times120=12\left(g\right)\)
\(\Rightarrow C\%_{MgSO_4}=\frac{12}{34,87}\times100\%=34,41\%\)
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