a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)

\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)

Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\left(1\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{56}{22,4}=2,5\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,5\left(mol\right)\\n_{C_2H_2}=1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,5.22,4}{33,6}.100\%\approx33,33\%\\\%V_{C_2H_2}\approx66,67\%\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=3,5\left(mol\right)\Rightarrow m_{O_2}=3,5.32=112\left(g\right)\)

Gọi số mol C, S là a, b

=> 12a + 32b = 7,68

PTHH: C + O₂ --t^o--> CO₂

_____a--------------->a

S + O₂ --t^o--> SO₂

b--------------->b

=> a + b = \(\dfrac{9,856}{22,4}=0,44\)

=> a = 0,32; b = 0,12

=> \(\left\{{}\begin{matrix}\%C=\dfrac{0,32.12}{7,68}.100\%=50\%\\\%S=\dfrac{0,12.32}{7,68}.100\%=50\%\end{matrix}\right.\)

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt : \(2Zn+O_2\underrightarrow{t^o}2ZnO|\)

         2         1        2

        a        0,5b    0,2

       \(2Mg+O_2\underrightarrow{t^o}2MgO|\)

          2        1           2

         b       0,5b       0,1

a) Gọi a là số mol của Zn

           b là số mol của Mg

\(m_{Zn}+m_{Mg}=15,4\left(g\right)\)

⇒ \(n_{Zn}.M_{Zn}+n_{Mg}.M_{Mg}=15,4g\)

  ⇒ 65a + 24b = 15,4g (1)

Theo phương trình : 0,5a + 0,5b = 0,15 (2)

            65a + 24b = 15,4g

            0,5a + 0,5b = 0,15

             ⇒ \(\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

         \(m_{Zn}=0,2.65=13\left(g\right)\)

         \(m_{Mg}=01.24=2,4\left(g\right)\)

b) Có : \(n_{ZnO}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

            ⇒ \(m_{ZnO}=0,2.81=16,2\left(g\right)\)

           \(n_{MgO}=\dfrac{0,1.2}{2}=0,1\left(mol\right)\)

           ⇒ \(m_{MgO}=0,1.40=4\left(g\right)\)

 Chúc bạn học tốt

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Mg}=b\left(mol\right)\left(a,b>0\right)\\ 2Zn+O_2\underrightarrow{^{to}}2ZnO\\a.........0,5a..........a\left(mol\right)\\ 2Mg+O_2\underrightarrow{^{to}}2MgO\\b..........0,5b...........b\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+24b=15,4\\0,5a+0,5b=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Mg}=0,1.24=2,4\left(g\right)\\ b.m_{ZnO}=81.0,2=16,2\left(g\right)\\ m_{MgO}=0,1.40=4\left(g\right)\)

Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=36\left(g\right)\)

\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)

PTHH: 

mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)

VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)

mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)

Đặt \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)

Theo đề: \(m_{hh}=39\left(g\right)\)

\(\Rightarrow m_{Al}+m_{Fe}=39\\ \Rightarrow27x+56y=39\left(1\right)\)

\(PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ \left(mol\right)....x\rightarrow..0.75x....0,5x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow..\dfrac{2}{3}y.....\dfrac{1}{3}y\)

Theo đề: \(n_{O_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)

\(\Rightarrow0,75x+\dfrac{2}{3}y=0,55\left(2\right)\)

\(\xrightarrow[\left(1\right)]{\left(2\right)}\left\{{}\begin{matrix}27x+56y=39\\0,75x+\dfrac{2}{3}y=0,55\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Fe}=0,6.56=33,6\left(g\right)\end{matrix}\right.\\ m_r=m_{Al_2O_3}+m_{Fe_3O_4}=0,5.0,2.102+\dfrac{1}{3}.0,6.232=56,6\left(g\right)\)

Gọi x, y lần lượt là số mol của Cu và Fe.

Ta có: \(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a. PTHH: 2Cu + O₂ ---t^o---> 2CuO (1)

3Fe + 2O₂ ---t^o---> Fe₃O₄ (2)

Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

Theo PT₍₁₎: \(n_{Cu}=n_{CuO}=0,2\left(mol\right)\)

Theo PT₍₁₎: \(n_{O_2}=\dfrac{1}{2}.n_{Cu}=\dfrac{1}{2}x\left(mol\right)\)

Theo PT₍₂₎: \(n_{O_2}=\dfrac{2}{3}.n_{Fe}=\dfrac{2}{3}y\left(mol\right)\)

=> \(\dfrac{1}{2}x+\dfrac{2}{3}y=0,3\)

Mà n_Cu = 0,2(mol)

Thay vào, ta được: \(\dfrac{1}{2}.0,2+\dfrac{2}{3}y=0,3\)

=> y = 0,3(mol)

=> \(m_{Cu}=0,2.64=12,8\left(g\right)\)

\(m_{Fe}=0,3.56=16,8\left(g\right)\)

b. \(\%_{Cu}=\dfrac{12,8}{12,8+16,8}.100\%=43,24\%\)

\(\%_{Fe}=100\%-43,24\%=56,76\%\)