Bạn tham khảo!

Ta có PTHH sau:

 $\text{2NaOH+CuCl2$\rightarrow$2NaCl+Cu(OH)2↓}$

Có $\text{nNaOH=CM.V=2.0,15=0,3 mol}$

Dựa vào PTHH) $\text{nCu(OH)2=$\dfrac{0,3}{2}$=0,15 mol}$

$\rightarrow$ $\text{a) mCu(OH)2=n.M=0,15.98=14,7g}$

b) Dựa vào PTHH có $\text{nCuCl2=nCu(OH)2=0,15 mol}$

$\rightarrow$ $\text{V CuCl2=$\dfrac{n}{CM}$=$\dfrac{0,15}{0,5}$=3 lít}$

Đổi $\text{3 lít}$ thành $\text{300ml}$

Có PTHH tiếp theo:

$\text{Cu(OH)2+H2SO4$\rightarrow$CuSO4+2H2O}$

Dựa vào PTHH) $\text{nH2SO4=nCu(OH)2=0,15 mol}$

$\rightarrow$ $\text{mH2SO4=n.M=0,15.98=14,7g}$

$\text{c) mdd H2SO4=$\dfrac{\text{mct.100}}{\text{C%}}$=$\dfrac{14,7.100}{9,8}$=150g}$

`#Ly`

$a)$

`150ml=0,15l`

`n_(NaOH` `=0,15.2=0,3(mol)`

PTHH: `2NaOH+CuCl_2->Cu(OH)_2+2NaCl`

Theo pt: `n_(Cu(OH)_2` `=(1)/(2)n_(NaOH` `=0,15(mol)`

`m_(Cu(OH)_2` `=0,15.98=14,7(g)`

$b)$

Theo pt: `n_(CuCl_2` `=` `(1)/(2)n_(NaOH` `=0,15(mol)`

`V_(ddCuCl_2` `=(0,15)/(0,5)=0,3(l)=300(ml)`

$c)$

PTHH: `Cu(OH)_2+H_2SO_4->CuSO_4+2H_2O`

Theo pt: `n_(H_2SO_4` `=n_(Cu(OH)_2` `=0,15(mol)`

`m_(H_2SO_4` `=0,15.98=14,7(g)`

`m_(ddH_2SO_4` `=(14,7.100)/(9,8)=150(g)`