In a test tube, 1.0mL of 0.10 M lead(II) nitrate and 10 mL of 1.0M sodium chloride are mixed. Calculate the theoretical mass of solid formed
Tbis is a limiting reagent (LR) problem.
Pb(NO3)2 + 2NaCl ==> PCl2 + 2NaNO3
mols Pb(NO3)2 = M x L = ?
mols NaCl = M x L = ?
Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols PbCl2.
Do the same and convert mols NaCl to mols PbCl2.
It is likely that the two numbers will not agree which means one of them is wrong; the correct value in LR problem is ALWAYS the smaller value and the reagent responsible for that number is the LR.
Using the smaller value, convert mols NaCl to g. g = mols x molar mass.
To calculate the theoretical mass of the solid formed, you need to determine the limiting reactant between lead(II) nitrate (Pb(NO3)2) and sodium chloride (NaCl). The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
First, let's write the balanced chemical equation for the reaction between lead(II) nitrate and sodium chloride:
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
According to the stoichiometry of the balanced equation, 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride to produce 1 mole of lead(II) chloride (PbCl2).
Now, let's calculate the number of moles of lead(II) nitrate and sodium chloride used in the reaction:
Moles of Pb(NO3)2 = Volume (in L) x Concentration (in mol/L)
= 1.0 mL x (1 L / 1000 mL) x 0.10 mol/L
= 0.010 mol
Moles of NaCl = Volume (in L) x Concentration (in mol/L)
= 10 mL x (1 L / 1000 mL) x 1.0 mol/L
= 0.100 mol
Now, we can determine the limiting reactant. Since 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride, we compare the moles of each reactant to find out which one is in excess.
The ratio of moles of Pb(NO3)2 to NaCl is 0.010 mol : 0.100 mol, or 1:10.
Since there is a 1:10 ratio, we can see that we have excess sodium chloride. Therefore, lead(II) nitrate is the limiting reactant.
Now, let's calculate the theoretical mass of solid lead(II) chloride formed:
Molar mass of PbCl2 = (207.2 g/mol for Pb) + (2 x 35.5 g/mol for Cl)
= 207.2 g/mol + 71.0 g/mol
= 278.2 g/mol
Theoretical mass of PbCl2 = Moles of Pb(NO3)2 x Molar mass of PbCl2
= 0.010 mol x 278.2 g/mol
= 2.782 g
Therefore, the theoretical mass of solid lead(II) chloride formed is 2.782 grams.