Question

Use the chemical equation and the table to answer the question.

Pb(NO3)2(aq) + 2KBr(aq) → PbBr2(s) + 2KNO3(aq)

Molar Mass (g/mol) of Reactants and Products Involved in a Reaction
Reactant or Product Molar Mass (g/mol)
Pb(NO3)2 331
KBr 119
PbBr2 367
KNO3 101
When 496.5 grams of Pb(NO3)2 reacts completely with KBr, how much will the total mass of the products be?

(1 point)
Responses

702.0 g
702.0 g

853.5 g
853.5 g

550.5 g
550.5 g

496.5 g

Answer 1

Use stoichiometry and conservation of mass.

Moles Pb(NO3)2 = 496.5 g / 331 g·mol−1 = 1.50 mol.
KBr needed = 2 mol KBr per mol Pb(NO3)2 → 1.50 × 2 = 3.00 mol KBr.
Mass KBr consumed = 3.00 mol × 119 g·mol−1 = 357.0 g.
Total mass of products = total mass of reactants = 496.5 + 357.0 = 853.5 g.

Answer: 853.5 g.