Final answer:
Produce 0.300 mole of potassium nitrate, 0.150 mole of lead(ii) nitrate would need to react, based on the stoichiometry of the balanced chemical equation which shows a 1:2 ratio between lead(ii) nitrate and potassium nitrate.
Step-by-step explanation:
The question asks, "How many moles of lead(ii) nitrate react to produce 0.300 mole of potassium nitrate according to the balanced chemical reaction?" To answer this, we need to examine the stoichiometry of the balanced chemical equation:
\ce {Pb(NO3)2 (aq) + 2 KI (aq) -> PbI2 (s) + 2 KNO3 (aq)}
From the equation, we can see that each mole of lead(ii) nitrate produces two moles of potassium nitrate. Therefore, to produce 0.300 mole of potassium nitrate, we'd need half of that amount of lead(ii) nitrate, since it has a 1:2 stoichiometric ratio with potassium nitrate. So the calculation is as follows:
0.300 mole KNO3 x (1 mole Pb(NO3)2 / 2 mole KNO3) = 0.150 mole Pb(NO3)2
Thus, to produce 0.300 mole of potassium nitrate, 0.150 mole of lead(ii) nitrate would need to react.