Bạn chia nhỏ câu hỏi ra được không:v, nhiều thế này mn ngại làm

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:

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

Đổi 2,5kg = 2500g

mC = 2500 . (100% - 16%) = 2100 (g)

nC = 2100/12 = 175 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 175 ---> 175 ---> 175

VO2 = 175 . 22,4 = 3920 (l)

mCO2 = 44 . 175 = 7700 (g)

\(m_C=\dfrac{2,5.\left(100-16\right)}{100}=2,1kg\)

\(m_C=2,1kg=2100g\)

\(n_C=\dfrac{m_C}{M_C}=\dfrac{2100}{12}=175mol\)

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

175  175           175    ( mol )

\(V_{O_2}=n_{O_2}.22,4=175.22,4=3920l\)

\(m_{CO_2}=n_{CO_2}.M_{CO_2}=175.44=7700g\)

Phần trăm khối lượng cacbon có trong than đá:

\(100\%-16\%=84\%\)

Khối lượng cacbon:

\(m_C=\dfrac{2,5\cdot84\%}{100\%}=2,1kg\Rightarrow n_C=175mol\)

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

175  175   175

\(V_{O_2}=175\cdot22,4=3920ml=3,92l\)

\(m_{CO_2}=175\cdot44=7700g=7,7kg\)

 C_{nguyên chất} = (100%-10%).1000 = 900 (g)

\(n_C=\dfrac{900}{12}=75\left(mol\right)\)

PTHH: C + O₂ ---t^o→ CO₂

Mol:    75    75

\(V_{O_2}=75.22,4=1680\left(l\right)\)

\(m_{CO_2}=75.44=3300\left(g\right)\)

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

\(n_C=\dfrac{7,5.96\%}{12}=0,6\left(mol\right)\\ C+O_2-^{t^o}\rightarrow CO_2\\ n_{O_2}=n_C=0,6\left(mol\right)\\ \Rightarrow V_{O_2}=0,6.22,4=13,44\left(l\right)\)

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

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

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

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

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

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

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

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

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

\(m_S=0,5\%.24=0,12\left(kg\right)=120\left(g\right)\\ \Rightarrow n_S=\dfrac{120}{32}=3,75\left(mol\right)\\ m_C=1,5\%.24=0,36\left(kg\right)=360\left(g\right)\\ \Rightarrow n_C=\dfrac{360}{12}=30\left(mol\right)\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ n_{SO_2}=n_S=3,75\left(mol\right)\\ \Rightarrow V_{SO_2\left(đktc\right)}=3,75.22,4=84\left(l\right)\\ n_{CO_2}=n_C=30\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=22,4.30=672\left(l\right)\)

(Chắc đề là 1,5% C)

24kg = 24000g

\(m_S=24000.0,5\%=120\left(g\right)\\ n_S=\dfrac{120}{32}=3,75\left(mol\right)\\ \%C=100\%-0,5\%-1,5\%=98\%\\ m_C=24000.98\%=23520\left(g\right)\\ n_C=\dfrac{23520}{12}=1960\left(mol\right)\)

PTHH:

      S + O₂ → (t^o) SO₂

3,75     3,75          3,75

      C + O₂  → (t^o) CO₂

   1960  1960         1960

\(V_{SO_2}=22,4.3,75=84\left(l\right)\\ V_{CO_2}=22,4.1960=43904\left(l\right)\)