Question

Answer 1

1 
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ b,Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ c,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ d,P_2O_5+3H_2O\rightarrow2H_3PO_4\) 
2 
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) 
           0,2                      0,2            0,2 
\(m_{ZnSO_4}=161.0,2=32,2\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:\dfrac{0,3}{1}>\dfrac{0,2}{1}\) 
=> CuO dư 
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}0,2\left(mol\right)\\ m_{CuO\left(p\text{ư}\right)}=\left(0,3-0,2\right).80=8\left(g\right)\\ m_{Cu}=0,2.64=12,8\left(g\right)\) 
\(m_{cr}=8+12,8=20,8\left(g\right)\)