`AgNO_3 + HCl -> AgCl \downarrow + HNO_3`
`0,15` `0,15` `0,15` `0,15` `(mol)`
`n_[AgNO_3]=[17.150]/[100.170]=0,15(mol)`
`a)C%_[HCl]=[0,15.36,5]/200 .100=2,7375%`
`b)C%_[HNO_3]=[0,15.63]/[150+200-0,15.143,5].100=2,88%`
\(n_{AgNO3}=\dfrac{17\%.150}{100\%.170}=0,15\left(mol\right)\)
Pt : \(AgNO_3+HCl\rightarrow AgCl+HNO_3|\)
0,15 0,15 0,15 0,15
a) \(n_{HCl}=\dfrac{0,15.1}{1}=015\left(mol\right)\)
⇒ \(m_{HCl}=0,15.36,5=5,475\left(g\right)\)
\(C_{ddHCl}=\dfrac{5,475.100}{200}=2,7375\%\)
b) \(n_{HNO3}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_{HNO3}=0,15.63=9,45\left(g\right)\)
\(m_{ddspu}=150+200-\left(0,15.143,5\right)=328,475\left(g\right)\)
\(C_{HNO3}=\dfrac{9,45.100}{328,475}=2,88\%\)
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