1) 800 mL of 0.025 M Pb(NO3)2 is mixed with 500 mL of 0.030 M K2CrO4. Calculate the grams of PbCrO4 that form.
I am confused on how to do this I got the answer it should be 4.85 g PbCrO4, but I would want to know how to get that. And also Limiting reactant is K2CrO4, but I don't know how to get limiting reactant from this kind of problem? Please show step by step how to do this! Thank you.
First set up the balanced equation:
K2CrO4 + Pb(NO3)2 -> PbCrO4 + 2KNO3
800ml of .025M solution is 0.02 moles
500ml of .030M solution is 0.015 moles
So, since you have fewer moles of K2CrO4, that will probably be the limiting reagent.
In fact, since 1 mole of the reactants produces 1 mole of PbCrO4, you will wind up with only .015 moles, having used up all the K2CrO4.
.015 moles of PbCrO4 = .015*(207.2+52+4*16) = 4.848 g
Thanks; I really appreciate it!
To calculate the mass of PbCrO4 formed in this reaction, you need to follow a few steps. Let's go through them one by one:
Step 1: Determine the limiting reactant.
The limiting reactant is the reactant that gets completely consumed during the reaction, thus limiting the amount of product formed. To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
Given:
Volume of Pb(NO3)2 solution = 800 mL = 0.800 L
Molarity of Pb(NO3)2 solution = 0.025 M
Volume of K2CrO4 solution = 500 mL = 0.500 L
Molarity of K2CrO4 solution = 0.030 M
First, convert the volumes (in liters) to the number of moles:
moles of Pb(NO3)2 = volume (L) × molarity = 0.800 L × 0.025 mol/L = 0.020 mol Pb(NO3)2
moles of K2CrO4 = volume (L) × molarity = 0.500 L × 0.030 mol/L = 0.015 mol K2CrO4
Next, you need to compare the stoichiometric ratio of the two reactants. From the balanced chemical equation:
2Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3
The ratio of Pb(NO3)2 to K2CrO4 is 2:1. Therefore, each mole of Pb(NO3)2 requires 1/2 mole of K2CrO4. To see which reactant is the limiting reactant, divide the moles of each reactant by their respective stoichiometric coefficients:
For Pb(NO3)2: 0.020 mol Pb(NO3)2 ÷ 2 = 0.010 mol K2CrO4
For K2CrO4: 0.015 mol K2CrO4 ÷ (1/2) = 0.030 mol Pb(NO3)2
Comparing these values, you can see that the 0.010 mol of K2CrO4 is the limiting reactant because it will fully react with the Pb(NO3)2.
Step 2: Use the limiting reactant to determine the moles of PbCrO4 formed.
Since the limiting reactant is K2CrO4, we should use the stoichiometry from the balanced equation to find the number of moles of PbCrO4 that will form.
According to the balanced equation:
2Pb(NO3)2 + K2CrO4 → PbCrO4 + 2KNO3
The stoichiometric ratio between K2CrO4 and PbCrO4 is 1:1. This means that for every mole of K2CrO4, we will form 1 mole of PbCrO4.
Therefore, the moles of PbCrO4 formed will be equal to the moles of the limiting reactant K2CrO4.
moles of PbCrO4 formed = 0.010 mol K2CrO4
Step 3: Calculate the mass of PbCrO4 formed.
To calculate the mass of PbCrO4 formed, you need to know its molar mass, which is the sum of the atomic masses of its elements.
PbCrO4:
Pb: atomic mass = 207.2 g/mol
Cr: atomic mass = 52.0 g/mol
O: atomic mass = 16.0 g/mol (there are four oxygen atoms in the formula)
molar mass of PbCrO4 = (207.2 g/mol) + (52.0 g/mol) + 4 × (16.0 g/mol) = 323.2 g/mol
Finally, calculate the mass of PbCrO4 formed using the moles and molar mass:
mass of PbCrO4 = moles of PbCrO4 × molar mass of PbCrO4
mass of PbCrO4 = 0.010 mol × 323.2 g/mol = 3.23 g PbCrO4
Therefore, the mass of PbCrO4 formed is approximately 3.23 grams.