Đáp án:

\(\begin{array}{l} a,\ a=m_{CuO}=9,32\ g.\\ b,\ V_{HCl}=105,646\ ml.\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:\\ K_2O+H_2O\to 2KOH\\ n_{KOH}=0,04\times 1=0,04\ mol.\\ Theo\ pt:\ n_{K_2O}=\dfrac{1}{2}n_{KOH}=0,02\ mol.\\ ⇒m_{K_2O}=0,02\times 94=1,88\ g.\\ ⇒a=m_{CuO}=11,2-1,88=9,32\ g.\\ b,\ PTHH:CuO+2HCl\to CuCl_2+H_2O\\ n_{CuO}=\dfrac{9,32}{80}=0,1165\ mol.\\ Theo\ pt:\ n_{HCl}=2n_{CuO}=0,233\ mol.\\ ⇒m_{HCl}=0,233\times 36,5=8,5045\ g.\\ ⇒m_{\text{dd HCl}}=\dfrac{8,5045}{7\%}=121,493\ g.\\ ⇒V_{HCl}=\dfrac{121,493}{1,15}=105,646\ ml.\end{array}\)

chúc bạn học tốt!

a)CuO+H2O→không xảy ra

K2O+H2O→2KOH

=>nKOH=0,04.1=0,04(mol)

Theo pt=>nK2O=$\frac{1}{2}$.nKOH=0,02(mol)

             =>nK2O=0,02.94=1,88(g)

=>a=mCuO=11,2-1,88=9,32(g)

b) nCuO=$\frac{9,32}{80}$=0,1165(mol)

PTHH:                CuO+2HCl→CuCl2+H2O

                       0,1165  0,233                            (mol)

=>mHCl=0,233.36,5=8,5045(g)

=>mdd HCl=$\frac{8,5045.100}{7}$=121,49(g)

=>Vdd HCl=$\frac{mdd HCl}{D}$=$\frac{121,49}{1,15}$=105,64 (ml)