Question

What quantity in moles of precipitate will be formed when 70.0 ml of 2.00 m nai is reacted with 100.0 ml of 0.900 m pb(no₃)₂ in the following chemical reaction? 2 nai (aq) pb(no₃)₂ (aq) → pbi₂ (s) 2 nano₃ (aq)

Answer 1

When 70.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂, 0.090 moles of PbI₂ precipitate will be formed, because Pb(NO₃)₂ is the limiting reactant.

Step-by-step explanation:

The question asks how much precipitate will be formed when 70.0 mL of 2.00 M NaI is mixed with 100.0 mL of 0.900 M Pb(NO₃)₂. To determine this, we will use stoichiometry. First, calculate the number of moles of each reactant:

• NaI: 70.0 mL × 2.00 M = 0.140 moles NaI.
• Pb(NO₃)₂: 100.0 mL × 0.900 M = 0.090 moles Pb(NO₃)₂.

Since the balanced equation tells us that 2 moles of NaI react with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbI₂, we can determine the limiting reactant. Pb(NO₃)₂ is limiting because it requires twice the amount of NaI to react completely, and we do not have double the moles of NaI compared to Pb(NO₃)₂. Thus, we will use the moles of Pb(NO₃)₂ to calculate the moles of PbI₂ precipitate formed.

The reactants are in a 2:1 ratio, so 0.090 moles of Pb(NO₃)₂ will produce 0.090 moles of PbI₂ according to the reaction stoichiometry.