32.12mL of 10.0 mol/l potassium chromate solution is reacted with 120.0m of 2.00mol/l lead (II) nitrate solution. The balanced equation is;
2KCrO4(aq)+Pb(NO3)2(aq)==>2KNO3(aq)+Pb(CrO4)2(s)
a. Determine the limiting reactant
b. What is the maximum mass, in grams, of lead(II) chromate that can be produced?
To determine the limiting reactant, we need to compare the amount of each reactant to the stoichiometry of the balanced equation.
a. Determine the limiting reactant:
1) Convert the volume of the potassium chromate solution to moles using the molarity:
Moles of KCrO4 = Volume (L) × Concentration (mol/L)
Moles of KCrO4 = (32.12 mL / 1000 mL/L) × 10.0 mol/L
Moles of KCrO4 = 0.3212 mol
2) Convert the volume of the lead (II) nitrate solution to moles using the molarity:
Moles of Pb(NO3)2 = Volume (L) × Concentration (mol/L)
Moles of Pb(NO3)2 = (120.0 mL / 1000 mL/L) × 2.00 mol/L
Moles of Pb(NO3)2 = 0.240 mol
3) Use the stoichiometry from the balanced equation to compare the moles of the reactants:
From the balanced equation: 2 moles of KCrO4 react with 1 mole of Pb(NO3)2
Therefore, the ratio of KCrO4 to Pb(NO3)2 is 2:1.
So, for every 2 moles of KCrO4, we need 1 mole of Pb(NO3)2.
Since the ratio of moles of KCrO4 to Pb(NO3)2 (0.3212 mol : 0.240 mol) is greater than 2:1, Pb(NO3)2 is the limiting reactant.
b. To find the maximum mass of lead(II) chromate that can be produced, we need to use the mole ratio from the balanced equation.
1) Calculate the moles of PbCrO4 formed using the amount of limiting reactant:
Moles of PbCrO4 = Moles of limiting reactant (Pb(NO3)2) × (1 mol PbCrO4 / 1 mol Pb(NO3)2)
Moles of PbCrO4 = 0.240 mol × (1 mol PbCrO4 / 1 mol Pb(NO3)2)
Moles of PbCrO4 = 0.240 mol
2) Calculate the maximum mass of PbCrO4 using its molar mass (given in grams per mole):
Mass of PbCrO4 = Moles of PbCrO4 × Molar mass of PbCrO4
Mass of PbCrO4 = 0.240 mol × (2 × (195.08 g/mol) + 4 × (51.9961 g/mol))
Mass of PbCrO4 = 0.240 mol × 323.16 g/mol
Mass of PbCrO4 = 77.6 g
Therefore, the maximum mass of lead(II) chromate that can be produced is 77.6 grams.
To determine the limiting reactant and the maximum mass of lead (II) chromate that can be produced, we need to compare the number of moles for each reactant.
Step 1: Calculate the number of moles for each reactant.
First, we calculate the number of moles for potassium chromate (K2CrO4):
Moles of K2CrO4 = volume (in L) x concentration (in mol/L)
Moles of K2CrO4 = (32.12 mL / 1000 mL/L) x 10.0 mol/L
Moles of K2CrO4 = 0.03212 L x 10.0 mol/L
Moles of K2CrO4 = 0.3212 mol
Next, we calculate the number of moles for lead (II) nitrate (Pb(NO3)2):
Moles of Pb(NO3)2 = volume (in L) x concentration (in mol/L)
Moles of Pb(NO3)2 = (120.0 mL / 1000 mL/L) x 2.00 mol/L
Moles of Pb(NO3)2 = 0.1200 L x 2.00 mol/L
Moles of Pb(NO3)2 = 0.2400 mol
Step 2: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed in the reaction and restricts the maximum amount of product formed. To determine the limiting reactant, we compare the mole ratios of the reactants in the balanced equation.
From the balanced equation: 2KCrO4(aq) + Pb(NO3)2(aq) ==> 2KNO3(aq) + Pb(CrO4)2(s)
The mole ratio between K2CrO4 and Pb(NO3)2 is 2:1.
Therefore, if the number of moles of K2CrO4 is double the number of moles of Pb(NO3)2, Pb(NO3)2 is the limiting reactant.
In this case, there are 0.3212 mol of K2CrO4 and 0.2400 mol of Pb(NO3)2.
Since the ratio between K2CrO4 and Pb(NO3)2 is approximately 1.34 (0.3212/0.2400), Pb(NO3)2 is the limiting reactant.
Step 3: Calculate the maximum mass of lead (II) chromate formed.
From the balanced equation, we know that each mole of Pb(NO3)2 reacts to produce 1 mole of Pb(CrO4)2.
The molar mass of Pb(CrO4)2 can be found by summing the atomic masses:
2 x atomic mass of Pb + 4 x atomic mass of Cr + 8 x atomic mass of O
= (2 x 207.2 g/mol) + (4 x 52.0 g/mol) + (8 x 16.0 g/mol)
= 504.4 g/mol
Therefore, the maximum mass of Pb(CrO4)2 formed can be calculated as follows:
Mass of Pb(CrO4)2 = moles of Pb(NO3)2 x molar mass of Pb(CrO4)2
= 0.2400 mol x 504.4 g/mol
= 120.96 g
So, the maximum mass of lead (II) chromate that can be produced is 120.96 grams.