Question

Answer 1

Bài 6.1:

\(n_{H_2SO_4}=\dfrac{122,5.40\%}{98}=0,5\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ a.n_{CuSO_4}=n_{CuO}=n_{H_2SO_4}=0,5\left(mol\right)\\ a.m_{CuSO_4}=160.0,5=80\left(g\right)\\ b.m_{CuO}=0,5.80=40\left(g\right)\)

Answer 2

Bài 6.2:

\(n_{Al}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ a.V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(mol\right)\\ b.n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\\ m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)