Đáp án:
\(\begin{array}{l}
b)\\
M:Kali(K)\\
c)\\
{M_{KCl}} = 14,9g\\
d)\\
{m_{{\rm{dd}}HCl}} = 36,5g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2M + 2HCl \to 2MCl + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
\Rightarrow {n_M} = 2{n_{{H_2}}} = 0,1 \times 2 = 0,2mol\\
\Rightarrow {M_M} = \dfrac{{7,8}}{{0,2}} = 39\\
\Rightarrow M:Kali(K)\\
c)\\
2K + 2HCl \to 2KCl + {H_2}\\
{n_{KCl}} = {n_K} = 0,2mol\\
\Rightarrow {M_{KCl}} = 0,2 \times 74,5 = 14,9g\\
d)\\
{n_{HCl}} = {n_K} = 0,2mol\\
{m_{HCl}} = 0,2 \times 36,5 = 7,3g\\
{m_{{\rm{dd}}HCl}} = \dfrac{{{m_{HCl}} \times 100}}{{20}} = \dfrac{{7,3 \times 100}}{{20}} = 36,5g
\end{array}\)