For the chemical reaction, 2 KI + Pb ( NO 3 ) 2 ⟶ PbI 2 + 2 KNO 3 what mass of lead(II) nitrate in grams is needed to produce 4.25 mol of potassium nitrate?

Question

For the chemical reaction, 

what mass of lead(II) nitrate in grams is needed to produce  of potassium nitrate?

Answer 1

2KI + Pb(NO₃)₂ ==> PbI₂ + 2KNO₃ ... balanced equation

1 mole of lead nitrate, Pb(NO₃)₂, produces 2 moles of potassium nitrate, KNO₃

Using this same mole ratio, we find how many moles of lead nitrate are needed to produce 4.25 moles KNO₃

4.25 moles KNO₃ x 1 mole Pb(NO₃)₂ / 2 moles KNO₃ = 2.125 moles Pb(NO₃)₂

Finally, we convert this to mass by using the molar mass of lead nitrate (331 g/mol):

2.125 moles Pb(NO₃)₂ x 331 g/mol = 703 g Pb(NO₃)₂ needed