Question

What quantity in moles of precipitate will be formed when 208.0 ml of 0.300 m lii is reacted with excess pb(no₃)₂ in the following chemical reaction? 2 lii(aq) pb(no₃)₂(aq) → pbi₂(s) 2 lino₃(aq)

Answer 1

To find the quantity of precipitate formed, we can use the stoichiometry of the balanced chemical equation. In this case, when 208.0 mL of 0.300 M LiI is reacted with excess Pb(NO3)2, 0.0312 moles of PbI2 precipitate will be formed.

Step-by-step explanation:

To determine the quantity of precipitate formed when 208.0 mL of 0.300 M LiI is reacted with excess Pb(NO3)2, we need to use the stoichiometry of the balanced chemical equation. From the equation, we can see that for every 2 moles of LiI, 1 mole of PbI2 precipitate is formed. Since the concentration of LiI is given in moles per liter (M), we need to convert the volume to liters and then use the molarity to find the number of moles of LiI. Finally, we can use the mole ratio to find the number of moles of PbI2 formed.

First, we convert the volume of LiI solution to liters:

208.0 mL x (1 L / 1000 mL) = 0.208 L

Next, we calculate the number of moles of LiI:

0.208 L x 0.300 M = 0.0624 moles of LiI

Finally, we use the mole ratio from the balanced equation:

0.0624 moles of LiI x (1 mole of PbI2 / 2 moles of LiI) = 0.0312 moles of PbI2