Question

5.8 g of an inorganic metal oxide ore mainly contains 90% of Nickel (III) oxide which decomposes to elemental nickel and oxygen gas. Produced 880 mL of oxygen were collected over water at a temperature of 25°C with a total pressure of 747 mmHg. What is the percent yield of decomposition reaction? (PH20:23.7 mmHg at 25°C)

Answer 1

5.8 g x 0.90 = 5.22 g = mass of the Ni2O3.
2Ni2O3 ==> 4Ni(s) + 3O2(g)
Use PV = nRT and solve for n = moles O2.
V = 880 mL = 0.880 L
R = 0.08206 L*atm/mol*K
T = 25 C = 298 K
Ptotal = 747 mm Hg = pO2 + pH2O = pO2 + 23.7. Solve for pO2 and convert to atm with atm = pO2 in mm/760 = 723.3/760 = = 0.9517 atm.
Plugging all of these numbers in you have
0.9517*0.880 = n*0.08206*298. Solve for n = moles O2 gas.
moles O2 gas x 32 g/mol = ? grams O2 gas. Then
% yield = grams O2 gas/5.22 g)*100 = ?
Post your work if you get stuck but I've done almost all of it for you.