\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.8....................................0.4\)
\(m_{Na}=0.8\cdot23=18.4\left(g\right)\)
\(m_{Cu}=37.6-18.4=19.2\left(g\right)\)
\(n_{CuO}=\dfrac{32.8}{80}=0.41\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(1...........1\)
\(0.41..........0.4\)
\(LTL:\) \(\dfrac{0.41}{1}>\dfrac{0.4}{1}\Rightarrow CuOdư\)
\(m_{Cu}=0.4\cdot64=25.6\left(g\right)\)
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