$1)$

Đặt $n_{CaCO_3}=x(mol)$

$PTHH:CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$

Theo PT: $n_{CO_2}=n_{CaCO_3}=x(mol)$

Ta có $m_{\text{rắn giảm}}=m_{CaCO_3}-m_{CaO}=22$

$\Rightarrow 100x-56x=22$

$\Rightarrow x=0,5(mol)$

$\Rightarrow m_{CaO}=0,5.56=28(g)$

$m_{CaCO_3(p/ứ)}=0,5.100=50(g)$

$2)$

$a)PTHH:Mg+2HCl\to MgCl_2+H_2(1)$

$MgO+2HCl\to MgCl_2+H_2O(2)$

$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$

Theo PT(1): $n_{Mg}=n_{H_2}=0,1(mol)$

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

$\Rightarrow m_{MgO}=6,4-2,4=4(g)$

$b)n_{MgO}=\dfrac{4}{40}=0,1(mol)$

Từ 2 PT: $\Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,2+0,2=0,4(mol)$

$\Rightarrow m_{HCl}=0,4.36,5=14,6(g)$

1)

Phản ứng xảy ra:

\(CaC{{\text{O}}_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)

Khối lượng rắn giảm là khối lượng của \(CO_2\)

\( \to {m_{C{O_2}}} = 22{\text{ gam}} \to {{\text{n}}_{C{O_2}}} = \frac{{22}}{{44}} = 0,5{\text{ mol = }}{{\text{n}}_{CaC{{\text{O}}_3}{\text{ phản ứng}}}} = {n_{CaO}}\)

\( \to {m_{CaO}} = 0,5.(40 + 16) = 28{\text{ gam}}\)

\({m_{CaC{{\text{O}}_3}{\text{ phản ứng}}}} = 0,5.100 = 50{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

\(MgO + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2O}\)

Ta có:

\({n_{{H_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol = }}{{\text{n}}_{Mg}}\)

\( \to {m_{Mg}} = 0,1.24 = 2,4{\text{ gam}} \to {{\text{m}}_{MgO}} = 6,4 - 2,4 = 4{\text{ gam}}\)

\( \to {n_{MgO}} = \frac{4}{{24 + 16}} = 0,1{\text{ mol}}\)

\( \to {n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,1.2 + 0,1.2 = 0,4{\text{ mol}}\)

\( \to {m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}}\)