Đáp án:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

a) Theo PTHH: \(n_{Al}=n_{AlCl_3}=2:2\)

\(\Rightarrow n_{AlCl_3}=n_{Al}.1=0,1\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

b) Theo PTHH: \(n_{Al}=n_{H_2}=2:3\)

\(\Rightarrow n_{H_2}=n_{Al}.\dfrac{3}{2}=0,1.\dfrac{3}{2}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36\left(l\right)\)

c) \(H_2+CuO\rightarrow\ Cu+H_2O\)

Theo PTHH: \(n_{H_2}=n_{CuO}=1:1\)

\(\Rightarrow n_{CuO}=n_{H_2}.1=0,15\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

Giải thích các bước giải:

nAl = 2,7 / 27 = 0,1 (mol)

2Al + 6HCl -> 2AlCl3 + 3H2

0,1                     0,1         0,15

mAlCl3 = 0,1 * 133,5 = 13,35 (gam)

VH2 = 0,15 * 22,4 = 3,36 (lít)

CuO + H2 -to-> Cu + H2O

0,15     0,15

mCuO = 0,15 * 80 = 12 (gam)