Bài 3.6
a)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,4--->0,8------>0,4--->0,4
mHCl = 0,8.36,5 = 29,2 (g)
\(m_{dd.HCl}=\dfrac{29,2.100}{14,6}=200\left(g\right)\)
\(V_{dd.HCl}=\dfrac{200}{1,115}=179,37\left(ml\right)\)
b) mdd sau pư = 26 + 200 - 0,4.2 = 225,2 (g)
\(C\%_{ZnCl_2}=\dfrac{0,4.136}{225,2}.100\%=24,16\%\)
Bài 3.7
a)
\(n_{H_2SO_4}=1,6.0,125=0,2\left(mol\right)\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2<-----0,2----->0,2
=> mZn = 0,2.65 = 13 (g)
=> mCu = 21 - 13 = 8 (g)
\(\left\{{}\begin{matrix}\%Cu=\dfrac{8}{21}.100\%=38,1\%\\\%Zn=\dfrac{13}{21}.100\%=61,9\%\end{matrix}\right.\)
b)
\(C_{M\left(ZnSO_4\right)}=\dfrac{0,2}{0,125}=1,6M\)
