Câu 2:

$a)n_{O_2}=\dfrac{9,6}{32}=0,3(mol)$

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

$S+O_2\xrightarrow{t^o}SO_2$

Theo PT: $n_C+n_S=0,3$

Mà $12n_C+32n_S=5,6$

$\Rightarrow n_C=0,2(mol);n_S=0,1(mol)$

$\Rightarrow m_C=0,2.12=2,4(g);m_S=0,1.32=3,2(g)$

$b)$ Theo PT: $n_{CO_2}=n_C=0,2(mol);n_{SO_2}=n_S=0,1(mol)$

$\Rightarrow \%V_{CO_2}=\dfrac{0,2}{0,1+0,2}.100\%\approx 66,67\%$

$\Rightarrow \%V_{SO_2}=100-66,67=33,33\%$ 

Lại có $m_{CO_2}=0,2.44=8,8(g);m_{SO_2}=0,1.64=6,4(g)$

$\Rightarrow \%m_{CO_2}=\dfrac{8,8}{8,8+6,4}.100\%\approx 57,89\%$

$\Rightarrow \%m_{SO_2}=100-57,89=42,11\%$

Câu 3:

$n_{O_2}=\dfrac{9,6}{32}=0,3(mol);n_{CO_2}=\dfrac{8,8}{44}=0,2(mol)$

$PTHH:2CO+O_2\xrightarrow{t^o}2CO_2(1)$

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

Theo PT(1): $n_{O_2(1)}=0,1(mol);n_{CO}=0,2(mol)$

$\Rightarrow n_{O_2(2)}=0,3-0,1=0,2(mol)$

Theo PT(2): $n_{H_2}=2n_{O_2(2)}=0,4(mol)$

$\Rightarrow m_{H_2}=0,4.2=0,8(g);m_{CO}=0,2.28=5,6(g)$

$\Rightarrow \%V_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%$

$\Rightarrow \%V_{H_2}=100-33,33=66,67\%$

$\%m_{CO}=\dfrac{5,6}{5,6+0,8}.100\%=87,5\%$

$\%m_{H_2}=100-87,5=12,5\%$

Câu 4:

$a)$ Đặt $n_{Al}=x(mol);n_{Fe}=y(mol)$

$\Rightarrow 27x+56y=55,2(1)$

$PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

Theo PT: $0,75x+\dfrac{2}{3}y=\dfrac{22,4}{22,4}=1(2)$

Từ $(1)(2)\Rightarrow x=0,8(mol);y=0,6(mol)$

$\Rightarrow n_{Al_2O_3}=0,4(mol);n_{Fe_3O_4}=0,2(mol)$

$\Rightarrow m_{Al_2O_3}=0,4.102=40,8(g);m_{Fe_3O_4}=0,2.232=46,4(g)$

$\Rightarrow \%m_{Al}=\dfrac{0,8.27}{55,2}.100\%=39,13\%$

$\Rightarrow \%m_{Fe}=100-39,13=60,87\%$

$\%m_{Al_2O_3}=\dfrac{40,8}{40,8+46,4}.100\%=46,79\%$

$\%m_{Fe_3O_4}=100-46,79=53,21\%$

$b)PTHH:2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

Theo PT: $n_{KMnO_4}=2n_{O_2}=2(mol)$

$\Rightarrow m_{KMnO_4}=2.158=316(g)$