a.

$n_{N_2}=\frac{6,72}{22,4}=0,3(mol)$ 

$m_{N_2}=0,3.28=8,4(g)$

b.

$n_{CO_2}=\frac{2,24}{22,4}=0,1(mol)$ 

$m_{CO_2}=0,1.44=4,4(g)$

c.

$n_{H_2}=\frac{8,96}{22,4}=0,4(mol)$ 

$m_{H_2}=0,4.2=0,8(g)$

d.

$n_{SO_2}=\frac{13,44}{22,4}=0,6(mol)$ 

$m_{S_2}=0,6.64=38,4(g)$

Đáp án:

\(\begin{array}{l} a,\ m_{N_2}=8,4\ g.\\ b,\ m_{CO_2}=4,4\ g.\\ c,\ m_{H_2}=0,8\ g.\\ d,\ m_{SO_2}=38,4\ g.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ n_{N_2}=\dfrac{6,72}{22,4}=0,3\ mol.\\ \Rightarrow m_{N_2}=0,3\times 28=8,4\ g.\\ b,\ n_{CO_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ \Rightarrow m_{CO_2}=0,1\times 44=4,4\ g.\\ c,\ n_{H_2}=\dfrac{8,96}{22,4}=0,4\ mol.\\ \Rightarrow m_{H_2}=0,4\times 2=0,8\ g.\\ d,\ n_{SO_2}=\dfrac{13,44}{22,4}=0,6\ mol.\\ \Rightarrow m_{SO_2}=0,6\times 64=38,4\ g.\end{array}\)

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