\(n_{H_2}=\dfrac{0,784}{22,4}=0,035\left(g\right)\)
Zn + 2HCl → ZnCl2 + H2
Fe + 2HCl → FeCl2 + H2
2Al + 6HCl → 2AlCl3 + 3H2
Theo 3 pt ta có: \(2n_{HCl}=n_{H_2}\Rightarrow n_{HCl}=\dfrac{1}{2}.0,035=0,0175\left(mol\right)\)
\(m_{HCl}=0,0175.36,5=0,63875\left(g\right)\)
Theo ĐLBTKL ta có:
\(m_{muối}=1,75+0,63875-0,035.2=2,38175\left(g\right)\)
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