$VD1:\\ a,\text{Đặt }n_{Al}=x(mol);n_{Mg}=y(mol)\\ \Rightarrow 27x+24y=7,8(1)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ PTHH:4Al+3O_2\xrightarrow{t^o}2Al_2O_3\\ 2Mg+O_2\xrightarrow{t^o}2MgO\\ \Rightarrow 0,75x+0,5y=0,2(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow m_{Al}=0,2.27=5,4(g);m_{Mg}=0,1.24=2,4(g)\\ b,n_{Al_2O_3}=0,5x=0,1(mol)\Rightarrow m_{Al_2O_3}=0,1.102=10,2(g)\\ n_{MgO}=y=0,1(mol)\Rightarrow m_{MgO}=0,1.40=4(g)\\ VD2:\\ a,\text{Đặt }n_{CH_4}=x(mol);n_{C_2H_6}=y(mol)\\ \Rightarrow x+y=\dfrac{6,72}{22,4}=0,3(1)\\ PTHH:CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O\\ 2C_2H_6+7O_2\xrightarrow{t^o}4CO_2+6H_2O\\ \Rightarrow n_{O_2}=2n_{CH_4}+3,5n_{C_2H_6}=2x+3,5y=\dfrac{24}{32}=0,75(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow m_{CH_4}=0,2.16=3,2(g);m_{C_2H_6}=0,1.30=3(g)\\ b,\Sigma n_{CO_2}=x+2y=0,2+0,2=0,4(mol)\\ \Rightarrow V_{CO_2}=0,4.22,4=8,96(l)$

a)

Vì dụ 1

gọi n_Al =a(mol) ; n_Mg = b(mol)

=> m_Al = 27a (g) và m_Mg = 24b(g)

mà m_{Al +} m_{Mg =} 7.8 (g)

=> 27a + 24b = 7.8

theo PT(1) => n_O2 = 3_/4 . n_Al = 3_/4 .a

theo PT (2) => n_O2 = 1_/2 . n_Mg = 1/2 b

mà tổng n_O2 = 0,2 (mol)

=> 3_/4 a + 1/2b = 0.2

Do đó: \(\left\{\begin{matrix}27a+24b=7.8\\\frac{3}{4}a+\frac{1}{2}b=0.2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a=0.2\left(mol\right)\\b=0.1\left(mol\right)\end{matrix}\right.\)

=> m_Al = 0.2 x 27=5.4(g)

=> m_Mg = 7.8 - 5.4 = 2.4(g)

b) Do đó: \(\left\{\begin{matrix}27a+24b=7.8\\\frac{3}{4}a+\frac{1}{2}b=0.2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a=0.2\left(mol\right)\\b=0.1\left(mol\right)\end{matrix}\right.\)