Đáp án + Giải thích các bước giải:

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

$b\big)$

$n_{H_2}=\frac{5,6}{22,4}=0,25(mol)$

Theo PT: $n_{O_2}=\frac{1}{2}n_{H_2}=0,125(mol)$

$\to m_{O_2}=0,125.32=4(g)$

$\to V_{O_2(đktc)}=0,125.22,4=2,8(l)$

$c\big)$

$2Na+2H_2O\to 2NaOH+H_2$

Theo PT: $n_{H_2}=\frac{1}{2}n_{H_2O}=n_{O_2}=0,125(mol)$

$\to V_{H_2(đktc)}=0,125.22,4=2,8(l)$

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

$\text{A.}$

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

$\text{B.}$

$\text{Ta có: $n_{H_{2}}$=$\dfrac{5,6}{22,4}$= 0,25 (mol)}$

$\text{Theo PT (1): $n_{O_{2}}$=$\dfrac{1}{2}$$n_{H_{2}}$=$\dfrac{1}{2}$.0,25= 0,125 (mol)}$

$\text{⇒ $V_{O_{2}}$=0,125.22,4= 2,8 (lít)}$

$\text{⇒ $m_{O_{2}}$=0,125.32= 4 (g)}$

$\text{C.}$

$\text{2Na + $2H_{2}O$→2NaOH + $H_{2}$ (2)}$

$\text{Theo PT (1): $n_{H_{2}O}$=$n_{H_{2}}$= 0,25 (mol)}$

$\text{Theo PT (2): $n_{H_{2}}$=$\dfrac{1}{2}$$n_{H_{2}O}$=$\dfrac{1}{2}$.0,25= 0,125 (mol)}$

$\text{⇒ $V_{H_{2}}$=0,125.22,4= 2,8 (lít)}$