Question

Prob 1. Lead(II) nitrate reacts with potassium iodide to form lead(II) iodide and potassium nitrate. How many g of lead(II) iodide can be made from 140 g of lead(II) nitrate and excess potassium iodide? How many g of potassium iodide will be used in the reaction?

Prob 2. Sodium sulfate and barium chloride react to form barium sulfate and sodium chloride. How many g of sodium sulfate are required to make 120 g of barium sulfate, assuming you have an excess of barium chloride?

Answer 1

Ans Prob 1:

Mass of Pb(NO₃)₂ = 140 g

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

No. of moles of Pb(NO₃)₂ = 140/331.2 = 0.4227 mol

Reaction: Pb(NO₃)₂ + 2 KI --> PbI₂ + 2 KNO₃

We can see from the reaction that 1 mol of Pb(NO₃)₂ gives 1 mol of PbI_2.

So, 0.4227 mol of Pb(NO₃)₂ gives 0.4227 mol of PbI_2.

Molar mass of PbI₂ = 461.01 g/mol

Mass of 0.4227 mol of PbI₂ = 0.4227 x 461.01 = 194.87 g

194.87 g of lead(II) iodide can be made from 140 g of lead(II) nitrate.

Reaction: Pb(NO₃)₂ + 2 KI --> PbI₂ + 2 KNO₃

We can see from the reaction that 1 mol of Pb(NO₃)₂ uses 2 mol KI.

So, 0.4227 mol of Pb(NO₃)₂ uses 0.8454 mol KI.

Molar mass of KI = 166.0028 g/mol

Mass of 0.8454 mol KI = 0.8454 x 166.0028 = 140.34 g

140.34 g of potassium iodide will be used in the reaction

Ans Prob 2:

Mass of BaSO₄ required = 120 g

Molar mass of BaSO₄ = 233.38 g/mol

No. of moles of BaSO₄ = 120/233.38 = 0.5142 mol

Reaction: Na₂SO₄ + BaCl₂ --> 2 NaCl + BaSO₄

We can see from the reaction that 1 mol of Na₂SO₄ gives 1 mol of BaSO₄

So, 0.5142 mol of BaSO₄ is formed from 0.5142 mol of Na₂SO_4.

Molar mass of Na₂SO₄ = 142.04 g/mol

Mass of 0.5142 mol of Na₂SO₄ = 0.5142 x 142.04 = 73.037 g

73.037 g of sodium sulfate are required to make 120 g of barium sulfate