Answer: 10.3 grams of Copper oxide is left unreacted.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} CuO=\frac{50g}{79.5g}=0.63moles[/tex]
The balanced chemical reaction is:
[tex]CuO+H_2SO_4\rightarrow CuSO_4+H_2O[/tex]
According to stoichiometry :
1 moles of [tex]H_2SO_4[/tex] require = 1 mole of [tex]CuO[/tex]
Thus 0.5 moles of [tex]H_2SO_4[/tex] will require=[tex]\frac{1}{1}\times 0.5=0.5moles[/tex] of [tex]CuO[/tex]
Thus [tex]H_2SO_4[/tex] is the limiting reagent as it limits the formation of product and [tex]CuO[/tex] is the excess reagent.
moles of CuO left = (0.63-0.5) mol= 0.13 mol
Mass of [tex]CuO[/tex] left =[tex]moles\times {\text {Molar mass}}=0.13moles\times 79.5g/mol=10.3g[/tex]
10.3 grams of Copper oxide is left unreacted.
