Question

c) tính m

Answer 1

a, PT: \(M+2HCl\rightarrow MCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{MCl_2}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow M_{MCl_2}=\dfrac{9,5}{0,1}=95\left(g/mol\right)\)

\(\Rightarrow M_M+35,5.2=95\Rightarrow M_M=24\left(g/mol\right)\)

→ M là Mg.

c, Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)