what is the equation?

Question

2 NaCl (aq) + Pb(NO3)2 (aq) → 2 NaNO3 (aq) + PbCl2 (s)

1. How many grams of lead (II) chloride are produced from the reaction of 10.3 g of NaCl and 50.8 g of Pb(NO3)2? 

What is the limiting reactant? 

Answer 1

Balanced Equation: 2 NaCl (aq) + Pb(NO₃)₂ (aq) → 2 NaNO₃ (aq) + PbCl₂ (s)

Relevant Info: Molar Mass of NaCl = 58.44 g/mol

Molar Mass of Pb(NO₃)₂ = 331.2 g/mol

Molar Mass of PbCl₂ = 278.1 g/mol

Stoichiometry: Set up for each reactant.

For NaCl: 10.3g NaCl × (1 mol NaCl / 58.44g NaCl) × (1 mol PbCl₂ / 2 mol NaCl) × (278.1 g PbCl₂ / 1 mol PbCl₂) = 24.51g PbCl₂

For Pb(NO₃)₂: 50.8g Pb(NO₃)₂ × (1 mol Pb(NO₃)₂ / 331.2g Pb(NO₃)₂) × (1 mol PbCl₂ / 1 mol Pb(NO₃)₂) × (278.1 g PbCl₂ / 1 mol PbCl₂) = 42.66g PbCl₂

Final Answer: Compare the two. The smaller one is the answer and the amount produced from the limiting reactant.

24.51g PbCl₂ is produced.

NaCl is the limiting reactant.