Câu 1:
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PT: \(n_{MgCl_2}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
c, \(m_{MgCl_2}=0,2.95=19\left(g\right)\)
d, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
Câu 2:
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)
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