a)

n C = (1000 - 1000.5%)/12 = 475/6 mol

C + O2 --to--> CO2

n O2 = n C = 475/6 mol

=> V O2 = 22,4 . 475/6 = 1773,33(lít)

b) V CO2 = V O2 = 1773,33(lít)

 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

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

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

Lượng than nguyên chất:

V O 2 = n O 2 .22,4 = 79,17.22,4 = 1773,4(l)

a/ trong 1 kg than có 5% tạp chất ko cháy => 95% khí cacbon cháy dc ( ko bik khoảng nì đúng ko >: mik ko chắc về cách trình bày chặng này bạn thông cảm )

=> Khối lượng cacbon có trong 1kg than:

\(1.95\%=0,95\)(kg) \(=950g\)

CTHH :\(C+O_2\rightarrow CO_2\)

             1  :    1      :    1

         475  :    475  : 475 

\(n_c=\frac{m_c}{M_c}=\frac{950}{2}=475\left(mol\right)\)

\(n_{O_2}=\frac{475.1}{1}=475\left(mol\right)\)

\(V_{O_2}=n_{O_2}.22,4=475.22,4=10640\left(l\right)\)

B/ \(n_{CO_2}=\frac{475.1}{1}=475\left(mol\right)\)

\(V_{CO_2}=n_{CO_2}.22,4=475.22,4=10640\left(l\right)\)

Bạn tham khảo tại đây nhé

https://sites.google.com/site/hoahocquan10/bai-tap/bai-tap-hoa-8/hoa-8-chuong-iv

m_C = 1000.95% = 950 (g)

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

Ta có: \(n_C=\dfrac{950}{12}=\dfrac{475}{6}\left(mol\right)\)

Theo PT: \(n_{O_2}=n_{CO_2}=n_C=\dfrac{475}{6}\left(mol\right)\)

a+b, \(V_{O_2}=V_{CO_2}=\dfrac{475}{6}.22,4\approx1773,33\left(l\right)\)

\(n_{CO_2}=\dfrac{25}{44}\left(kmol\right)\)

PTHH: C + O₂ --t^o--> CO₂

         \(\dfrac{25}{44}\)<---------------\(\dfrac{25}{44}\)

\(n_C=\dfrac{\dfrac{25}{44}}{85\%}=\dfrac{125}{187}\left(kmol\right)\\ m_{than}=\dfrac{\dfrac{125}{187}.12}{100\%-7\%}=8,625\left(kg\right)\)

Đổi 1kg = 1000g

mC = 1000 . (100% - 10%) = 900 (g)

nC = 900/12 = 75 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 75 ---> 75

VO2 = 75 . 22,4 . 5 = 8400 (l)

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:

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