\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1.......0,2..............0,1.........0,1
a, \(V=n_{HCl}.M=0,2.1=0,2\left(l\right)\)
b, \(m_{dd}=20.1,1=22\left(g\right)\)
\(m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
\(m_{H_2}=2.0,1=0,2\left(g\right)\)
=>\(m_{dd_{sau}}=m_{dd}+m_{Zn}-m_{H_2}=22+6,5-0,2=28,3\left(g\right)\)
\(C\%=\dfrac{13,6}{28,3}.100\%=48\%\)
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