Question

Answer 1

a) nhôm sủi bọt và sinh ra khí H2 
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
            0,1      0,15                                 0,15 
\(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 
           0,15       0,15
\(m_{CuO}=0,15.80=12\left(g\right)\)

Answer 2

2Al+3H2SO4->Al2(SO4)3+3H2

0,1---0,15------------------------------0,15

H2+CuO-to>Cu+H2O

0,15----0,15

n Al=0,1 mol

=>VH2=0,15.24,79=3,7185l

=>a=0,15.98=14,7g

=>m CuO=0,15.80=12g