Cu + O2 \(\underrightarrow{to}\) CuO (1)
4Al + 3O2 \(\underrightarrow{to}\) 2Al2O3 (2)
Gọi \(n_{Cu}=n_{Al}=x\left(mol\right)\)
Theo pT1: \(n_{CuO}=n_{Cu}=x\left(mol\right)\)
\(\Rightarrow m_{CuO}=80x\left(g\right)\)
Theo Pt2: \(n_{Al_2O_3}=\frac{1}{2}n_{Al}=0,5x\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,5x\times102=51x\left(g\right)\)
Ta có: \(m_{CuO}+m_{Al_2O_3}=13,1\)
\(\Leftrightarrow80x+51x=13,1\)
\(\Leftrightarrow131x=13,1\)
\(\Leftrightarrow x=0,1\)
Vậy \(n_{Cu}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\times64=6,4\left(g\right)\)
\(\Rightarrow m_{Al}=0,1\times27=2,7\left(g\right)\)
Vậy \(m=m_{Cu}+m_{Al}=6,4+2,7=9,1\left(g\right)\)
Đặt :
nCu = nAl= x mol
Cu + 1/2O2 -to-> CuO
x______________x
2Al + 3/2O2 -to-> Al2O3
x_______________0.5x
mY = 80x + 0.5x* 102 = 13.1
=> x = 0.1
mhh= m = 0.1*64 + 0.1*27 = 9.1g
