1.\(n_{CuO}=\frac{8}{80}=0,1mol\)
PTHH:
CuO + H2SO4 -> CuSO4 + H2O
Theo pt: \(n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,1mol\)
\(m_{dd_{H_2SO_4}}=\frac{0,1.98.100}{24,5}=40g\)
\(m_{ddX}=8+40=48g\)
\(m_{CuSO_4}=0,1.160=16g\) \(C\%=\frac{m_{ct}}{m_{dd}}.100=\frac{16}{48}.100=33,33\%\)
CuO + H2SO4 -> CuSO4 + H2O
Theo pt: \(n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,1mol\)
\(m_{dd_{H_2SO_4}}=\frac{0,1.98.100}{24,5}=40g\)
\(m_{ddX}=8+40=48g\)
\(m_{CuSO_4}=0,1.160=16g\) \(C\%=\frac{m_{ct}}{m_{dd}}.100=\frac{16}{48}.100=33,33\%\)
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