\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)
\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
\(m_{ddHCl}=\dfrac{21,9.100}{7,3}=300\left(g\right)\)
\(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,3.136=40,8\left(g\right)\)
\(m_{ddspu}=19,5+300-\left(0,3.2\right)=318,9\left(g\right)\)
\(C_{ZnCl2}=\dfrac{40,8.100}{318,9}=12,8\)0/0
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