Đáp án:

Giải thích các bước giải:

$a)$ $m_{C}=1,5.80\%=1,2(kg)$

$n_C=\dfrac{1,2}{12}=0,1(kmol)$

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

Theo pt: $n_{O_2}=n_C=0,1(kmol)$

$V_{O_2}=0,1.22,4=2,24(m^3)=2240(l)$ 

$b)$

$2C_4H_{10}+13O_2\xrightarrow{t^o}8CO_2+10H_2O$

Theo pt: $V_{O_2}=\dfrac{13}{2}V_{C_4H_{10}}=32,5(l)$

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

$\text{A.}$

$\text{C + $O_{2}$$\xrightarrow{t^o}$$CO_{2}$}$

$\text{Ta có: $m_{C}$=$\dfrac{1,5 × 80}{100}$= 1,2 (Kg)= 12000 (g)}$

$\text{⇒ $n_{C}$=$\dfrac{m_{C}}{M_{C}}$=$\dfrac{12000}{12}$= 100 (mol)}$

$\text{Theo PT ta có: $n_{O_{2}}$=$n_{C}$= 100 (mol)}$

$\text{⇒ $V_{O_{2}}$=$n_{O_{2}}$ × 22,4=100 × 22,4= 2240 (lít)}$

$\text{B.}$

$\text{$2C_{4}H_{10}$ + $13O_{2}$$\xrightarrow{t^o}$$8CO_{2}$ + $10H_{2}O$}$

$\text{Theo PT ta có: $V_{O_{2}}$=$\dfrac{13}{2}$$V_{C_{4}H_{10}}$=$\dfrac{13}{2}$ × 5= 32,5 (lít)}$