If 15.0mL of 2.50M NaCl is mixed with 60.0mL of 0.300M Pb(NO3)2, what is the mass of the precipitate
You know this is a limiting reagent because amounts are given for BOTH reactants.
Here is a worked example of a limiting reagent problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To find the mass of the precipitate formed in the reaction between NaCl and Pb(NO3)2, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in a chemical reaction, thus determining the maximum amount of product that can be formed.
First, let's calculate the number of moles of NaCl and Pb(NO3)2:
For NaCl:
Volume = 15.0 mL
Concentration = 2.50 M
Convert mL to L: 15.0 mL / 1000 = 0.015 L
Moles of NaCl = concentration * volume = 2.50 M * 0.015 L = 0.0375 moles
For Pb(NO3)2:
Volume = 60.0 mL
Concentration = 0.300 M
Convert mL to L: 60.0 mL / 1000 = 0.060 L
Moles of Pb(NO3)2 = concentration * volume = 0.300 M * 0.060 L = 0.018 moles
Next, we need to determine the stoichiometry of the reaction. From the balanced chemical equation, we know that the molar ratio between NaCl and Pb(NO3)2 is 2:1. This means that for every 2 moles of NaCl, 1 mole of Pb(NO3)2 is consumed.
Since the mole ratio is 2:1, Pb(NO3)2 is the limiting reactant because it has fewer moles (0.018 moles) compared to NaCl (0.0375 moles).
To find the mass of the precipitate, we need to use the stoichiometry of the balanced equation. The molar mass of the precipitate can be obtained from the formula of the product, which is PbCl2.
The molar mass of PbCl2:
Pb = 207.2 g/mol
Cl = 35.5 g/mol x 2 = 71.0 g/mol
Molar mass of PbCl2 = 207.2 g/mol + 71.0 g/mol = 278.2 g/mol
Now, we can determine the mass of the precipitate formed:
Mass = moles of Pb(NO3)2 * molar mass of PbCl2 = 0.018 moles * 278.2 g/mol = 5.0076 grams
Therefore, the mass of the precipitate formed is approximately 5.0076 grams.