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

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

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

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

PTHH: C + O₂ --t^o--> CO₂

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

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

a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,4---------------->0,4

=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)

b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH₄ dư, O₂ hết

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

                      0,4-------->0,2

=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(m_C=1000\cdot90\%=900\left(kg\right)\)

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

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

\(75............75\)

\(m_{CO_2}=75\cdot44=3300\left(kg\right)\)

\(m_{CO_2\left(tt\right)}=3300\cdot80\%=2640\left(kg\right)=2.64\left(\text{tấn}\right)\)

Lượng than nguyên chất:

n C O 2 = n O 2 = 79,17(mol) → V C O 2 = V O 2  = 1773,4(l)

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)

TN1: Gọi (n_Cu, n_Al, n_Fe) = (a,b,c)

=> 64a + 27b + 56c = 14,3 (1)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             b----------------------->1,5b

            Fe + 2HCl --> FeCl₂ + H₂

              c----------------------->c

=> 1,5b + c = 0,3 (2)

TN2: Gọi (n_Cu, n_Al, n_Fe) = (ak,bk,ck)

=> ak + bk + ck = 0,6 (3)

\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

            ak--->0,5ak

            4Al + 3O₂ --t^o--> 2Al₂O₃

           bk--->0,75bk

            3Fe + 2O₂ --t^o--> Fe₃O₄

           ck-->\(\dfrac{2}{3}ck\)

=> 0,5ak + 0,75bk + \(\dfrac{2}{3}ck\) = 0,4 (4)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,1\left(mol\right)\\c=0,15\left(mol\right)\\k=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{14,3}.100\%=22,38\%\\\%m_{Al}=\dfrac{0,1.27}{14,3}.100\%=18,88\%\\\%m_{Fe}=\dfrac{0,15.56}{14,3}.100\%=58,74\%\end{matrix}\right.\)

\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)

\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)