Bài giải+ giải thích

$1)$ Tính khối lượng

$a)  2,479l O_2 (đktc)$

$n_{O_2 (đkc)} = \dfrac{V}{24,79} = \dfrac{2,479}{24,79} = 0,1 (mol)$

$m_{O_2 (đkc)} = n . M = 0,1 . (16.2) = 3,2 (g)$

$b)  19,832l CO_2 (đktc)$

$n_{CO_2 (đkc)}= \dfrac{V}{24,79} = \dfrac{19,832}{24,79} = 0,8 (mol)$

$m_{CO_2 (đkc)} = n . M = 0,8 . (12+16.2)= 35,2 (g)$

$2)$ Tính thể tích 

$a) 4g H_2$

$n_{H_2}= \dfrac{m}{M} = \dfrac{4}{2} = 2(mol)$

$V_{H_2 (đkc)} = n . 24,79 = 2 . 24,79 = 49,58 (l)$

$b) 3,2g SO_2$

$n_{SO_2} = \dfrac{m}{M} = \dfrac{3,2}{64} = 0,05 (mol)$

$V_{SO_2 (đkc)} = n . 24,79 = 0,05 . 24,79 = 1,2395 (l)$

$\huge\text{Cho mình câu tlhn}$

Em tham khảo nha:

\(\begin{array}{l}
1)\\
a)\\
{n_{{O_2}}} = \dfrac{V}{{24,79}} = \dfrac{{2,479}}{{24,79}} = 0,1\,mol\\
{m_{{O_2}}} = n \times M = 0,1 \times 32 = 3,2g\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{24,79}} = \dfrac{{19,832}}{{24,79}} = 0,8mol\\
{m_{C{O_2}}} = n \times M = 0,8 \times 44 = 35,2g\\
2)\\
a)\\
{n_{{H_2}}} = \dfrac{m}{M} = \dfrac{4}{2} = 2\,mol\\
{V_{{H_2}}} = n \times 24,79 = 2 \times 24,79 = 49,48l\\
b)\\
{n_{S{O_2}}} = \dfrac{m}{M} = \dfrac{{3,2}}{{64}} = 0,05\,mol\\
{V_{S{O_2}}} = n \times 24,79 = 0,05 \times 24,79 = 1,2395l
\end{array}\)