Answer the following questions based on this reaction:
Pb(NO3)2 (aq) + 2 NaCl (aq) --> PbCl2 (s) + 2 NaNO3 (aq)
a) If 225 mL of 12.95 M Pb(NO3)2(aq) are reacted with a solution made with 5.05 g of NaCl (aq), how many grams of lead (II) chloride will be precipitated
b) If the actual yield of lead (II) chloride is 1.06 g, what is the percent yield?
PLEASE HELP :( I HAVE MY FINAL IN A COUPLE WEEKS!
This is another LR problem. Look at the last problem to see the process used. For mols of solutions, mols = M x L = ? while that of solids is mols = grams/molar mass. When you find the mass PbCl2 that is the theoretical yield.
Then yield is
% yield = (actual yield/theoretical yield)*100 = ?
Post your work if you get stuck.
a) To calculate the grams of lead (II) chloride precipitated, we need to find out how many moles of PbCl2 are formed first. Then we can convert moles to grams using the molar mass.
Step 1: Calculate the moles of Pb(NO3)2 reacted:
Moles of Pb(NO3)2 = volume (L) x concentration (M)
Given:
Volume = 225 mL = 0.225 L
Concentration = 12.95 M
Moles of Pb(NO3)2 = 0.225 L x 12.95 M = 2.91 moles
Step 2: Calculate the moles of PbCl2 precipitated:
From the balanced equation, we can see that the mole ratio between Pb(NO3)2 and PbCl2 is 1:1. Thus, the moles of PbCl2 formed will also be 2.91 moles.
Step 3: Calculate the grams of PbCl2:
Grams of PbCl2 = moles x molar mass
The molar mass of PbCl2 = (207.2 g/mol) + 2(35.45 g/mol) = 278.1 g/mol
Grams of PbCl2 = 2.91 moles x 278.1 g/mol = 809.07 g
Therefore, approximately 809.07 grams of lead (II) chloride will be precipitated.
b) To calculate the percent yield, we need to compare the actual yield with the theoretical yield.
Given:
Actual yield = 1.06 g
Theoretical yield = 809.07 g (calculated in part a)
Step 1: Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (1.06 g / 809.07 g) x 100 = 0.131% (rounded to three decimal places)
Therefore, the percent yield is approximately 0.131%.