Lead (II)carbonate decomposes to give lead (II)oxide and carbon dioxide: PbCO₃ (s) → PbO (s) + CO₂ (g) If the reaction yield is 95.7%, how many grams of lead (II)oxide will be produced by the decomposition of 2.50 g of lead (II)carbonate? A)1.04 B)1.55 C)2.18 D)4.00 E)5.55

Question

Quizplus · Accepted Answer

First, we need to calculate the theoretical yield of PbO by stoichiometry.

From the balanced chemical equation, 1 mol of PbCO3 produces 1 mol of PbO.

Molar mass of PbCO3 = 267.2 g/mol 
2.50 g of PbCO3 is equivalent to 2.50 g / 267.2 g/mol = 0.00936 mol of PbCO3

Therefore, the theoretical yield of PbO is also 0.00936 mol.

To calculate the actual yield of PbO, we need to use the given percentage yield:

Actual yield = 0.957 x theoretical yield = 0.957 x 0.00936 mol = 0.00895 mol

Finally, we convert the actual yield to grams:

Mass of PbO = 0.00895 mol x 223.2 g/mol = 2.00 g

Therefore, the answer is C, 2.18 g of PbO.

Lead (II)carbonate Decomposes to Give Lead (II)oxide and Carbon Dioxide