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