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a. `-` `2H_2 + O_2 \overset{t^o}\to 2H_2O`

`-` $n_{H}$ `= ` $\dfrac{44,8}{22,4}$ `= 2` `(mol)`

`->` $n_{H_2}$ `= 2 × 2 = 4` `(mol)`

`-` $n_{O}$ `= ` $\dfrac{22,4}{22,4}$ `= 1` `(mol)`

`->` $n_{O_2}$ `= 1 × 2 = 2` `(mol)`

`->` $m_{H}$ `= 4 × 1 = 4` `(g)`

`->` $m_{O}$ `= 2 × 16 = 32` `(mol)`

`->` $m_{H}$` : `$m_{O}$ `= 4 : 32` 

`->` `H` và `O` có tỉ lệ `1 : 8` hay `4 : 32`

b. `-` `%m_H =` $\dfrac{4}{36}$`× 100% = 11,11 %` 

`-` `%m_O =` $\dfrac{32}{36}$`× 100% = 88,89%`

Đáp án:

\({m_H}:{m_O}  = 1:8\)

\(\% {m_H} = 11,11\% ; \% {m_O} = 88,89\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(2{H_2} + {O_2}\xrightarrow{{{t^o}}}2{H_2}O\)

Ta có:

\({n_{{H_2}}} = \frac{{44,8}}{{22,4}} = 2{\text{ mol}} \to {{\text{n}}_H} = 2{n_{{H_2}}} = 4{\text{ mol}}\)

\({n_{{O_2}}} = \frac{{22,4}}{{22,4}} = 1{\text{ mol}} \to {{\text{n}}_O} = 2{n_{{O_2}}} = 2{\text{ mol}}\)

\( \to {m_H} = 4.1 = 4{\text{ gam;}}{{\text{m}}_O} = 2.16 = 32{\text{ gam}}\)

\( \to {m_H}:{m_O} = 4:32 = 1:8\)

\(\% {m_H} = \frac{4}{{4 + 32}} = 11,11\%  \to \% {m_O} = 88,89\% \)