Đáp án + Giải thích các bước giải:

 `Cu(OH)_2+2HCl->CuCl_2+2H_2O`

$n_{HCl}$`=`$\dfrac{9,125}{36,5}$`=0,25` mol

`a)`

Theo pt:$n_{Cu(OH)_{2}}$`=0,25÷2=0,125` mol

$n_{CuCl_{2}}$`=0,25÷2=0,125` mol

$m_{Cu(OH)_{2}}$`=0,125.98=12,25` g

`b)`

$m_{CuCl_{2}}$`=0,125.135=16,875` g

Em tham khảo!

Ta có PTHH sau:

$\text{Cu(OH)2+2HCl$\rightarrow$CuCl2+2H2O}$

$\text{Có nHCl=$\frac{m}{M}$=$\frac{9,125}{36,5}$=0,25 mol}$

$\text{Có nCu(OH)2=$\frac{0,25}{2}$=0,125 mol}$

$\rightarrow$ $\text{a) mCu(OH)2=n.M=0,125.98=12,25g}$

$\text{Có nCuCl2=nCu(OH)2=0,125 mol}$
$\rightarrow$ $\text{b) mCuCl2=n.M=0,125.135=16,875g}$

#TRANPHAMGIABAO