Question

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Answer 1

\(n_{CuO}=\dfrac{8}{80}=0,1mol\\ n_{H_2SO_4}=\dfrac{100.24,5\%}{100\%.98}=0,25mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,25}{1}\Rightarrow H_2SO_4.dư\\ n_{CuSO_4}=n_{H_2SO_4.pư}=n_{CuO}=0,1mol\\ C_{\%CuSO_4}=\dfrac{0,1.160}{8+100}\cdot100\%\approx14,81\%\\ C_{\%H_2SO_4.dư}=\dfrac{\left(0,25-0,1\right)98}{8+100}\cdot100\%\approx18,61\%\)