Đáp án + Giải thích các bước giải:

$a\big)Mg+2HCl\to MgCl_2+H_2$

$b\big)$

$n_{Mg}=\frac{4,8}{24}=0,2(mol)$

Theo PT: $n_{H_2}=n_{Mg}=0,2(mol)$

$\to V_{H_2(đkc)}=0,2.24,79=4,958(l)$

$c\big)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{H_2O}=n_{H_2}=0,2(mol)$

$\to m_{H_2O(thực\,tế)}=0,2.(100\%-5\%).18=3,42(g)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b)\\
{n_{Mg}} = \dfrac{{4,8}}{{24}} = 0,2mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,2mol\\
{V_{{H_2}}} = 0,2 \times 24,79 = 4,958l\\
c)\\
2{H_2} + {O_2} \to 2{H_2}O\\
{n_{{H_2}O}} = {n_{{H_2}}} = 0,2mol\\
{m_{{H_2}O}} = 0,2 \times 18 = 3,6g\\
{m_{{H_2}{O_{td}}}} = 3,6 - \dfrac{{3,6 \times 5}}{{100}} = 3,42g
\end{array}\)