Answer:
The options e and d are correct.
Explanation:
Mass of NiO = 7.5 g
Moles of NiO = ![\frac{7.5 g}{74.69 g/mol}=0.10 mol](https://tex.z-dn.net/?f=%5Cfrac%7B7.5%20g%7D%7B74.69%20g%2Fmol%7D%3D0.10%20mol)
Moles of sulfuric acid = n
Volume of sulfuric acid ,V= 50 mL = 0.050 L
Molarity of sulfuric acid ,M = 6 mol/L
![n=M\time V=6mol/L\times 0.050 L =0.3 mol](https://tex.z-dn.net/?f=n%3DM%5Ctime%20V%3D6mol%2FL%5Ctimes%200.050%20L%20%3D0.3%20mol)
![NiO + H_2SO_4\rightarrow NiSO_4 + H_2O](https://tex.z-dn.net/?f=NiO%20%2B%20H_2SO_4%5Crightarrow%20NiSO_4%20%2B%20H_2O)
According to reaction, 1 mole of NiO reacts with 1 mole of sulfuric acid.
Then 0.10 moles of NiO reacts with :
of sulfuric acid.
As we can see that sulfuric acid is in excess amount, so the amount of the product will depend upon amount of NiO.
According to reaction, 1 mole of NiO gives with 1 mole of
.
Then 0.10 moles of NiO wil give :
of
.
Molar mass of
= 154.75 g/mol
Mass of 0.10 moles of
:
= 154.75 g/mol × 0.10 mol = 15.475 g
Theoretical mass of
= 15.475 g
Experimental yield of
= 17.4 g
Percentage yield :
![\Yield=\frac{\text{Experimental mass}}{\text{Theoretical mass}}\times 100](https://tex.z-dn.net/?f=%5CYield%3D%5Cfrac%7B%5Ctext%7BExperimental%20mass%7D%7D%7B%5Ctext%7BTheoretical%20mass%7D%7D%5Ctimes%20100)
Percentage yield of
:
![\Yield=\frac{17.4}{15.475 g}\times 100=112\%](https://tex.z-dn.net/?f=%5CYield%3D%5Cfrac%7B17.4%7D%7B15.475%20g%7D%5Ctimes%20100%3D112%5C%25)
Moles of
= 262.85 g/mol × 0.10 mol = 26.285 g
Experimental yield of
= 17.4 g
Percentage yield of
:
![\Yield=\frac{17.4}{26.285 g}\times 100=66.2\%](https://tex.z-dn.net/?f=%5CYield%3D%5Cfrac%7B17.4%7D%7B26.285%20g%7D%5Ctimes%20100%3D66.2%5C%25)