Đổi: 24kg = 24000g

24kg than đá có chứa 0,5% tạp chất lưu huỳnh và 1,5% tạp chất khác không cháy được

⇒ nS = 120 / 32 = 3,75 mol

mS= 0,5% . 24=0,12(kg)=120(g)

->nS= 120/32=3,75(mol)

mC=(100% - 2%). 24=23,52(kg)=23520(g)

-> nC= 23520/12=1960(mol)

PTHH: S + O2 -to-> SO2

3,75______________3,75(mol)

C + O2 -to-> CO2 

1960______1960(mol)

=> V(SO2,đktc)=3,75 x 22,4=84(l)

V(CO2,đktc)= 1960 x 22,4= 43904(l)

PT: \(C+O_2\underrightarrow{t^o}CO_2\) (1)

\(S+O_2\underrightarrow{t^o}SO_2\) (2)

Ta có: m_S = 24.0,5% = 0,12 (kg) = 120 (g) ⇒ n_S = 120/32 = 3,75 (mol)

Theo PT (2): \(n_{SO_2}=n_S=3,75\left(mol\right)\)

\(\Rightarrow V_{SO_2}=3,75.22,4=84\left(l\right)\)

Ta có: m_C = 24 - 0,12 - 24.1,5% = 23,52 (kg) = 23520 (g)

\(\Rightarrow n_C=\dfrac{23520}{12}=1960\left(mol\right)\)

Theo PT (1): \(n_{CO_2}=n_C=1960\left(mol\right)\)

\(\Rightarrow V_{CO_2}=1960.22,4=43904\left(l\right)\)

$m_S = 24.0,5\% = 0,12(kg)$

$m_C = 24.(100\% - 0,5\% - 1,5\%) = 23,52(kg)$

$\Rightarrow n_S = 0,00375(kmol) = 3,75(mol)$

$n_C = 1,96(kmol) = 1960(mol)$

$C + O_2 \xrightarrow{t^o} CO_2$
$S + O_2 \xrightarrow{t^o} SO_2$

Theo PTHH : 

$n_{CO_2} = n_C = 1960(mol) ; n_{SO_2} = n_S = 3,75(mol)$

$\Rightarrow V_{CO_2} = 1960.22,4 = 43904(lít)$
$\Rightarrow V_{SO_2} = 3,75.22,4 = 84(lít)$

\(m_C=12\cdot\left(100-1.5-0.5\right)\%=11.76\left(kg\right)\)

\(n_C=\dfrac{11.76}{12}=0.98\left(kmol\right)\)

\(m_S=12\cdot0.5=6\left(kg\right)\)

\(n_S=\dfrac{6}{32}=0.1875\left(kmol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(C+O_2\underrightarrow{t^0}CO_2\)

\(V_{O_2}=\left(0.1875+0.98\right)\cdot22.4=26.152\left(kl\right)=26125\left(l\right)\)

0,5g = 0,0005 kg

\(m_C=36-0,0005-\left(36.1,5\%\right)=35,4595kg=35459,5g\)

\(n_C=\dfrac{35459,5}{12}=2954,95mol\)

\(n_S=\dfrac{0,5}{32}=0,015625mol\)

\(C+O_2\rightarrow\left(t^o\right)CO_2\)

2954,5            2954,5      ( mol )

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

\(0,015625\)         \(0,015625\)         ( mol )

\(V_{CO_2}=2954,5.22,4=66180,8l\)

\(V_{SO_2}=\)\(0,015625.22,4=0,35l\)

0,55 hay 0,55% mà 0,55 của cái gì em?

Chắc là : 0.5% tạp chất lưu huỳnh nhỉ 

\(\%C=100-0.5-1.5=98\%\)

\(m_C=\dfrac{24\cdot98}{100}=23.52\left(kg\right)\)

\(n_C=\dfrac{23.52\cdot10^3}{12}=1960\left(mol\right)\)

\(m_S=\dfrac{24\cdot0.5}{100}=0.12\left(kg\right)\)

\(n_S=\dfrac{0.12\cdot10^3}{32}=3.75\left(mol\right)\)

\(C+O_2\underrightarrow{t^0}CO_2\)

\(1960.......1960\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(3.75.......3.75\)

\(V_{CO_2}=1960\cdot22.4=43904\left(l\right)\)

\(V_{SO_2}=3.75\cdot22.4=84\left(l\right)\)

B1:

\(n_C=\dfrac{96\%.14.1000}{12}=1120\left(mol\right)\\ n_S=\dfrac{2,56\%.14.1000}{32}=11,2\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ n_{SO_2}=n_S=11,2\left(mol\right)\\ n_{CO_2}=n_C=1120\left(mol\right)\\ V_{CO_2\left(đktc\right)}=1120.22,4=25088\left(l\right)\\ n_{SO_2\left(đktc\right)}=11,2.22,4=250,88\left(l\right)\)

B2:

\(n_{C_2H_6}=\dfrac{1,8.\left(100\%-2\%\right).1000}{22,4}=78,75\left(mol\right)\\ C_2H_6+\dfrac{7}{2}O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\\ n_{O_2\left(đktc\right)}=\dfrac{7}{2}.78,75=275,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=275,625.22,4=6174\left(l\right)=6,174\left(m^3\right)\)