Question

Answer 1

a,\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: R + 2HCl → RCl₂ + H₂

Mol:    0,2    0,4        0,2         0,2

\(M_R=\dfrac{13}{0,2}=65\left(g/mol\right)\)

 ⇒ R là kim loại kẽm (Zn)

b,\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(m_{ddHCl}=\dfrac{14,6.100}{20}=73\left(g\right)\)

\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100}{73+13-0,2.2}=31,78\%\)

Answer 2

\(R + 2HCl ---> RCl_2 + H_2\)

\(n_{H_2}= \dfrac{4,48}{22,4}= 0,2 mol\)

\(Theo PTHH: n_R = n_{H_2} = 0,2 mol\)

\(\Rightarrow M_R= \dfrac{13}{0,2} = 65 ( Zn)\)

\(b)\)

\(n_{HCl}=2n_{H_2}= 0,4 mol\)\(\Rightarrow m_{HCl} = 14,6g\)

\(m_{dd HCl} =\dfrac{14,6 . 100}{20}= 73g\)

\(BTKL:\)

\(m_{dd ZnCl_2}= m_{Zn} + m_{dd HCl} - m_{H_2}\)

              \(= 13 + 73 - 0,2 . 2\)

              \(= 85,6 g\)

\(Theo PTHH:\)

\(m_{ZnCl_2}= 0,2 . 136= 27,2g\)

C%= \(\dfrac{27,2}{85,6} . 100 \) ~ 31,78 %