Question

Giúp mình câu 8 với câu 10 ik

Answer 1

\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
pthh : \(Zn+2HCl->ZnCl_2+H_2\)
           0,05                                  0,05 
\(n_{CuO}=\dfrac{6}{80}=0,075\left(mol\right)\)
pthh : \(H_2+CuO-t^o->Cu+H_2O\)
 LTL : 
\(\dfrac{0,05}{1}< \dfrac{0,075}{1}\)
=> CuO dư 
theo pthh : nCu = nH2 =0,05 (mol)
=> \(m_{Cu}=0,05.64=3,2\left(g\right)\)
theo pthh : \(n_{CuO\left(p\text{ư}\right)}=n_{H_2}=0,05\left(mol\right)\)
=> \(n_{CuO\left(d\right)}=0,075-0,05=0,025\left(mol\right)\)
=>\(m_{CuO}=0,025.80=2\left(g\right)\)

Answer 2

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

          0,1          0,15             0,05           0,15

PbO + H₂ --to--> Pb + H₂O

0,15    0,15

\(\rightarrow\left\{{}\begin{matrix}a=0,15.98=14,7\left(g\right)\\V=0,15.22,4=3,36\left(l\right)\\m_{PbO}=0,15.233=34,95\left(g\right)\end{matrix}\right.\)

Answer 3

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
pthh : \(4Al+6H_2SO_4->2Al_2\left(SO_4\right)_3+3H_2\)
           0,1      0,6                                      0,3 
=> \(V=V_{H_2}=0,3.22,4=6,72\) (L) 
=> a = \(m_{H_2SO_4}=0,6.98=58,8\left(g\right)\)
pthh : \(PbO+H_2->H_2O+Pb\)
          0,3       0,3 
=> \(m_{PbO}=0,3.217=65,1\left(g\right)\)
 

Answer 4

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(pthh:2Al+3H_2SO_4->Al_2\left(SO_{\text{ 4}}\right)_3+3H2\)
           0,1      0,15                                   0,15 
=> \(V=V_{H_2}=0,15.22,4=3,36\left(L\right)\)
=>\(a=m_{H_2so_4}=0,15.98=14,7\left(g\right)\)
\(pthh:PbO+H_2-t^o->Pb+H_2O\)
          0,15    0,15 
=> \(m_{PbO}=217.0,15=32,55\left(G\right)\)