Em tham khảo nha:

\(\begin{array}{l}
a)\\
2{H_2} + {O_2} \to 2{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{{O_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
\dfrac{{0,15}}{2} > \dfrac{{0,05}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}O}} = 2{n_{{O_2}}} = 0,1mol\\
{m_{{H_2}O}} = 0,1 \times 18 = 1,8g\\
b)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = \dfrac{2}{3}{n_{{O_2}}} = \frac{1}{{30}}mol\\
{m_{KCl{O_3}}} = \dfrac{1}{{30}} \times 122,5 = 4,0833g
\end{array}\)

$\text{Đáp án+Giải thích các bước giải:

$\text{$2H_{2}$ + $O_{2}$$\xrightarrow{t^o}$$2H_{2}O$ (1)}$

$\text{$2KClO_{3}$$\xrightarrow{t^o}$2KCl + $3O_{2}$ (2)}$

$\text{A.}$

$\text{Ta có: $n_{H_{2}}$=$\dfrac{3,36}{22,4}$= 0,15 (mol)}$

$\text{$n_{O_{2}}$=$\dfrac{1,12}{22,4}$= 0,05 (mol)}$

$\text{Lập tỉ lệ: $\dfrac{0,15}{2}$>$\dfrac{0,05}{1}$}$

$\text{⇒ $H_{2}$ dư, bài toán tính theo số mol của $O_{2}$}$

$\text{Theo PT (1) ta có: $n_{H_{2}O}$=2$n_{O_{2}}$=2 × 0,05= 0,1 (mol)}$

$\text{⇒ $m_{H_{2}O}$=0,1 × 18= 1,8 (g)}$

$\text{B.}$

$\text{Ta có: $n_{O_{2}}$= 0,05 (mol)}$

$\text{Theo PT (2) ta có: $n_{KClO_{3}}$=$\dfrac{2}{3}$$n_{O_{2}}$=$\dfrac{2}{3}$ × 0,05= $\dfrac{1}{30}$ (mol)}$

$\text{⇒ $m_{KClO_{3}}$=$\dfrac{1}{30}$ × 122,5≈ 4,08 (g)}$