Question

Answer 1

a) \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PTHH: \(n_{Mg}=n_{MgCl_2}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m=m_{Mg}=0,5.24=12\left(g\right)\)

c) Theo PTHH: \(n_{HCl}=2n_{H_2}=1\left(mol\right)\)

\(\Rightarrow C_{M\left(HCl\right)}=\dfrac{1}{0,15}=\dfrac{20}{3}M\)

d) \(m_{ddHCl}=150.1=150\left(g\right)\)

\(\Rightarrow m_{dd.sau.p\text{ư}}=150+12-0,5.2=161\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{161}.100\%=29,5\%\)