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Question

What quantity in moles of precipitate will be formed when 100.0 mL of 2.00 M NaI is reacted with 80.0 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

Stoichiometric calculation with quantities of both reactants implies a limiting reactant problem

Moles of NaI = .1 liters * (2 moles NaI/liter) = .2 moles

Moles of LN (lead nitrate) = .080 liters*(.9 moles LN/liter) = .072 moles

You need a ratio of NaI/LN of 2/1 and you have .2/.072 = 2.8 (NaI in excess, LN Limiting)

Use LN quantity to determine PbI₂

.072 moles LN(1 PbI₂/LN) = .072 PbI₂

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