chemistry question I would really love if someone can help me

Question

2.      How many grams of sodium nitrate can be produced with 60 grams of sodium chloride and 55 grams of lead IV nitrate? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction?

2 NaCl + Pb(NO₃)₂  ---> PbCl₂ + 2 NaNO₃

Answer 1

1). Write a correctly balanced equation:

2 NaCl + Pb(NO₃)₂  ---> PbCl₂ + 2 NaNO₃ (NOTE: this is NOT lead IV nitrate but is lead II nitrate)**

2). One easy way to identify the limiting reactant is to divide the moles of each reactant by the coefficient of the reactant in the balanced equation.

For NaCl we have 60 g NaCl x 1 mol / 58.4 g = 1.03 moles NaCl (÷2=0.515)

For Pb(NO₃)₂ we have 55 g x 1 mol / 331 g = 0.166 moles Pb(NO₃)₂ (÷1=0.166)

Pb(NO3)2 is the limiting reactant since 0.166 is less than 0.515

3). Find grams of NaNO₃ that can be produced:

Use the moles of limiting reactant and the mole ratio in the balanced equation.

0.166 mols Pb(NO₃)₂ x 2 mols NaNO₃ / mol Pb(NO₃)2 x 85 g/mol NaNO₃ = 28.2 g NaNO₃

4). To find grams of excess reactant (NaCl) left over, first find moles of NaCl used up, then subtract that from initial moles of NaCl present, and finally convert to grams of NaCl:

moles NaCl used up: 0.166 mols Pb(NO₃)₂ x 2 mols NaCl / mol Pb(NO₃)₂ = 0.332 mols NaCl used

moles NaCl left over = 1.03 mols - 0.332 mols = 0.698 mols NaCl left

grams NaCl left over = 0.698 mols x 58.4 g/mol = 40.8 g NaCl left

** If you really meant the question to relate to lead IV nitrate, you'd have to redo the problem using the molar mass of Pb(NO₃)₄ and write a newly balanced equation.