Question

How many grams of lead product would theoretically be produced between 17. 0 g potassium iodide, ki, and 25. 0 g of lead (ii) nitrate, pb(no3)2?.

Answer 1

The number of grams of lead product that would theoretically be produced from the given reaction is; Mass of lead product = 23.6 g

Potassium iodide reacting with Lead(II) Nitrate would yield the balanced equation;

2KI + Pb(NO₃)₂ = 2KNO₃ + PbI₂

2moles : 1 mole = 2 moles : 1 mole

From online tables;

Molar mass of KI = 166 g/mol

Molar mass of Pb(NO₃)₂ = 331.2 g/mol

We are given;

Mass of KI = 17 g

Mass of Pb(NO₃)₂ = 25 g

Thus;

Number of moles of PbI₂ yielded in order to find the limiting reactant will be;

no of moles of PbI₂ from KI = (17g/166 g/mol) × (1 mole of PbI₂/2 mol of KI)

no of moles of PbI₂ from KI = 0.0512 mols

Similarly;

no of moles of PbI₂ from Pb(NO₃)₂ = (25g/331.2 g/mol) × (1 mole of PbI₂/1 mol of Pb(NO₃)₂)

no of moles of PbI₂ from Pb(NO₃)₂ = 0.0755 moles

We can see that KI produced the least amount of moles. Thus;

Mass = 0.0512 * molar mass of  PbI₂

From tables, molar mass of  PbI₂ = 461 g/mol

Thus;

Mass of lead product = 461 * 0.0512

Mass of lead product = 23.6 g

Read more about theoretical yield at; brainly.com/question/21091465