2Al + 2NaOH +6H₂O→2Na[Al(OH)₄]+3H₂
1 mol H₂...................22.4 L
x moli......................3.36 L
x=3.36/24=0.14 moli H₂
2 moli Al............3 moli H₂
y moli Al...........0.14 moli H₂
y=0.14*2/3=0.09 moli Al
n=m/M
m=n*M=0.09*27=2.43 g Al
8.1 g aliaj...........2.43 g Al...........5.67g Cu
100 g aliaj..........x.......................y
x=100*2.43/8.1=30%Al
y=100*5.67/8.1=100-30=70%Cu