An excess lead nitrate solution was added to a 6.5 g sample mixture containing strontium nitrate and potassium bromide to precipitate out 0.8993 g of lead (II) bromide.

Question

Solve for the percent by mass of the individual components of the mixture.

Answer 1

Pb(NO₃)₂(aq) + Sr(NO₃)₂(aq) + 2KBr(aq) ==> PbBr₂(s) + Sr²⁺(aq) + NO₃^-(aq) + K⁺(aq)

molar mass PbBr₂ = 367 g/mol

moles PbBr₂ = 0.8993 g x 1 mol / 367 g = 0.00245 mols PbBr₂

moles Pb(NO₃)₂ present = 0.00245 mols (based on 1:1 mole ratio in the balanced equation)

mass Pb(NO₃)₂ = 0.00245 mols x 331 g/mol = 0.811 g Pb(NO₃)₂

moles KBr = 0.00245 mols PbBr₂ x 2 mol KBr / mol PbBr₂ = 0.0049 moles KBr

mass KBr = 0.0049 moles x 119 g = 0.583 g KBr

mass of Sr(NO₃)₂ = 6.5 g - 0.811 g - 0.583 g = 5.1 g Sr(NO₃)₂

Percent by mass:

For Pb(NO₃)₂: 0.811 g / 6.5 g (x100%) = 13%

For Sr(NO₃)₂: 5.1 g / 6.5 g (x100%) = 79%

For KBr: 0.583 g / 6.5 g (x100%) = 9%