CuO +H2SO4---->CuSO4 +H2O
Ta có
n\(_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
n\(_{H2SO4}=\frac{11,76.25}{100.98}=0,03\left(mol\right)\)
=>H2SO4 dư
Theo pthh
n\(_{H2SO4}=n_{CuO}=0,02\left(mol\right)\)
n\(_{H2SO4}dư=0,03-0,02=0,01\left(mol\right)\)
%m\(_{H2SO4}=\frac{0,01.98}{1,6+11,76}.100\%=7,34\%\)
Theo pthh
n\(_{CuSO4}=n_{CuO}=0,02\left(mol\right)\)
C%=\(\frac{0,02.160}{13,16}.100\%=24,32\%\)