Em tham khảo nha :

\(\begin{array}{l}
Ca + 2HCl \to CaC{l_2} + {H_2}\\
{n_{Ca}} = \dfrac{{2,4}}{{40}} = 0,06mol\\
\dfrac{{0,06}}{1} > \dfrac{{0,06}}{2} \Rightarrow Ca\text{ dư}\\
{n_{C{a_d}}} = 0,06 - \dfrac{{0,06}}{2} = 0,03mol\\
{m_{C{a_d}}} = 0,03 \times 40 = 1,2g\\
{n_{CaC{l_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,03mol\\
{m_{CaC{l_2}}} = 0,03 \times 111 = 3,33g
\end{array}\)

\(\begin{array}{l}
{n_{{H_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,03mol\\
{m_{{H_2}}} = 0,03 \times 2 = 0,06g
\end{array}\)

Đáp án + Giải thích các bước giải:

$\text{$n_{Ca}$ = $\dfrac{m_{Ca}}{M_{Ca}}$ = $\dfrac{2,4}{40}$ = 0,06 (mol)}$

$\text{PT: Ca + 2HCl $\rightarrow$ $CaCl_{2}$ + $H_{2}$}$

$\text{(mol) 0,03 $\leftarrow$ 0,06 $\rightarrow$ 0,03 $\rightarrow$ 0,03}$

$\text{Ta có: $\dfrac{n_{Ca}}{1}$ > $\dfrac{n_{HCl}}{2}$ (0,06 > 0,03)}$

$\text{$\Rightarrow$ Ca dư, HCl hết}$

$\text{$\Rightarrow$ PT tính theo số mol HCl}$

$\text{Theo PT ta có: $n_{Ca_{pứ}}$ = $\dfrac{1}{2}$$n_{HCl}$ = $\dfrac{1}{2}$ . 0,06 = 0,03 (mol)}$

$\text{$\Rightarrow$ $n_{Ca_{dư}}$ = $n_{Ca_{bđ}}$ - $n_{Ca_{pứ}}$ = 0,06 - 0,03 = 0,03 (mol)}$

$\text{$\Rightarrow$ $m_{Ca_{dư}}$ = $n_{Ca_{dư}}$ . $M_{Ca}$ = 0,03 . 40 = 1,2 (g)}$

$\text{Theo PT ta có: $n_{CaCl_{2}}$ = $\dfrac{1}{2}$$n_{HCl}$ = $\dfrac{1}{2}$ . 0,06 = 0,03 (mol)}$

$\text{$\Rightarrow$ $m_{CaCl_{2}}$ = $n_{CaCl_{2}}$ . $M_{CaCl_{2}}$ = 0,03 . (40 + 35,5.2) = 0,03 . 111 = 3,33 (g)}$

$\text{Theo PT ta có: $n_{H_{2}}$ = $\dfrac{1}{2}$$n_{HCl}$ = $\dfrac{1}{2}$ . 0,06 = 0,03 (mol)}$

$\text{$\Rightarrow$ $m_{H_{2}}$ = $n_{H_{2}}$ . $M_{H_{2}}$ = 0,03 . 2 = 0,06 (g)}$

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