\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,05 0,1 0,05
b) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
\(n_{CuCl2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
⇒ \(m_{CuCl2}=0,05.135=6,75\left(g\right)\)
\(m_{ddspu}=4+36,5=40,5\left(g\right)\)
\(C_{CuCl2}=\dfrac{6,75.100}{40,5}=16,67\)0/0
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