PTHH: $\text{C + $O_2$ $\xrightarrow{t^o}$ $CO_2$}$

a, Ta có: $\text{$n_{O_{2(p.ư)}}$ = $\dfrac{3,2}{32}$ = 0,1 mol}$

Theo PTHH $\rightarrow$ $\text{$n_{CO_2}$ = $n_{O_{2(p.ư)}}$ = 0,1 mol}$

$\rightarrow$ $\text{$m_{CO_2}$ = 0,1. 44 = 4,4 gam}$

b, Ta có: $\text{$n_{C_{(p.ư)}}$ = 0,3 mol}$

Theo PTHH $\rightarrow$ $\text{$n_{CO_2}$ = $n_{C_{(p.ư)}}$ = 0,3 mol}$

$\rightarrow$ $\text{$m_{CO_2}$ = 0,3. 44 = 13,2 gam}$

c, Ta có: $\text{$n_C$ = $\dfrac{6}{12}$ = 0,5 mol}$

             $\text{$n_{O_2}$ = $\dfrac{33,6}{22,4}$ = 1,5 mol}$

Ta có tỷ lệ:

     $\text{$\dfrac{n_C}{1}$ = $\dfrac{0,5}{1}$ < $\dfrac{n_{O_2}}{1}$ = $\dfrac{1,5}{1}$}$

$\rightarrow$ $\text{$O_2$ dư, C hết $\rightarrow$ Tính theo C}$

Theo PTHH $\rightarrow$ $\text{$n_{CO_2}$ = $n_C$ = 0,5 mol}$

$\rightarrow$ $\text{$m_{CO_2}$ = 0,5. 44 = 22 gam}$

$\text{#Kudo}$

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