a. PTHH: CuO + H2SO4 ---> CuSO4 + H2O
b. Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{CuO}=0,1\left(mol\right)\)
=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=4,9\%\)
=> \(m_{dd_{H_2SO_4}}=200\left(g\right)\)
c. Ta có: \(m_{dd_{CuSO_4}}=200+8=208\left(g\right)\)
Theo PT: \(n_{CuSO_4}=n_{CuO}=0,1\left(mol\right)\)
=> \(m_{CuSO_4}=0,1.160=16\left(g\right)\)
=> \(C_{\%_{CuSO_4}}=\dfrac{16}{208}.100\%=7,69\%\)
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