Answer:
The options e and d are correct.
Explanation:
Mass of NiO = 7.5 g
Moles of NiO = 
Moles of sulfuric acid = n
Volume of sulfuric acid ,V= 50 mL = 0.050 L
Molarity of sulfuric acid ,M = 6 mol/L


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 :

Percentage yield of
:

Moles of
= 262.85 g/mol × 0.10 mol = 26.285 g
Experimental yield of
= 17.4 g
Percentage yield of
:
