Theo đề bài ta có : \(\left\{{}\begin{matrix}nH2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\nCu=\dfrac{5,76}{64}=0,09\left(mol\right)\end{matrix}\right.\)
Ta có PTHH :
\(CuO+H2-^{t0}\rightarrow Cu+H2O\)
Ta có : nCu = 0,09 mol < nH2 = 0,1 mol
=> H = \(\dfrac{n\left(ch\text{ất}-thi\text{ếu}\right)}{n\left(ch\text{ất}-d\text{ư}\right)}.100\%=\dfrac{0,09}{0,1}.100\%=90\%\)
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