Bài 4 :
\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)
Pt : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
a) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{120}\simeq12,17\)0/0
c) \(m_{ddspu}=16,2+120=136,2\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{136,2}=19,97\)0/0
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