Stoichometry Chemistry Question About Concentration Ions

Question

You react 100.0 mL of 0.100 M lead(II) nitrate with 200.0 mL of 0.250 M potassium iodide. What is the concentration of I⁻ ions left in solution after the reaction is complete? 

 0 M 

 0.0667 M 

 0.100 M 

 0.250 M 

 0.167 M

Answer 1

Write the correctly balanced equation:

Pb(NO₃)₂(aq) + 2KI(aq) ==> 2KNO₃(aq) + PbI₂(s)

Find which reactant, if any, is in limiting supply:

moles Pb(NO₃)₂ present = 0.1 L x 0.100 mol/L = 0.01 moles

moles KI present = 0.200 L x 0.250 mol/L = 0.05 moles

Limiting reactant = Pb(NO₃)₂ because it takes 2 mol KI per mole Pb(NO₃)₂ and there is enough KI

Determine how many moles of KI will react with the 0.01 moles of Pb(NO₃)₂:

0.01 moles Pb(NO₃)₂ x 2 moles KI/mole Pb(NO₃)₂ = 0.02 moles KI will react

Determine moles KI remaining unreacted:

0.05 moles KI to start with less the 0.02 moles KI that reacts = 0.03 moles KI left over

Determine the final concentration of I^- ions:

Final volume of solution = 100.0 ml + 200.0 ml = 300.0 ml = 0.3000 L

0.03 moles KI/0.3000 L = 0.100 mol/L = 0.100 mol/L I^- = 0.100 M I^-