\(a,n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,05---->0,1------>0,1
\(\rightarrow x=C\%_{HCl}=\dfrac{0,1.36,5}{200}.100\%=1,825\%\)
\(b,\) PTHH: \(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
0,05----->0,1-------->0,05----------->0,1
\(\rightarrow m_{ddNaOH}=\dfrac{0,1.40}{10\%}=40\left(g\right)\\ \rightarrow m_{dd\left(sau.pư\right)}=40+200+4-0,05.98=239,1\left(g\right)\)
\(\rightarrow C\%_{NaCl}=\dfrac{0,1.58,5}{239,1}.100\%=2,45\%\)
\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\
pthh:CuO+HCl\rightarrow CuCl_2+H_2O\)
0,05 0,05 0,05 0,05
\(x=C\%_{HCl}=\dfrac{0,05.36,5}{200}.100\%=0,9125\%\\
pthh:CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)