In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?
mols K2CrO4 = M x L = ?
Using the coefficients in the balanced equation, convert mols K2CrO4 to mols PbCrO4(hint: 1 mol K2CrO4 = 1 mol PbCrO4)
Then convert mols PbCrO4 formed to grams by g = mols x molar mass = ?
In the reaction K2CrO4(aq) + PbCl2(aq) → 2KCl(aq) + PbCrO4(s), how many grams of PbCrO4 will precipitate out from the reaction between 500.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?
To find the grams of PbCrO4 that will precipitate out, we need to calculate the moles of PbCrO4 using the given information.
Step 1: Calculate the moles of K2CrO4:
To find the moles of K2CrO4, we can use the formula:
moles = Molarity x Volume (in liters)
Given: Volume of K2CrO4 = 500.0 milliliters = 500.0 / 1000 = 0.5 liters
Molarity of K2CrO4 = 3.0 M
moles of K2CrO4 = 3.0 M x 0.5 L
moles of K2CrO4 = 1.5 moles
Step 2: Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
From the balanced equation, we can see that the stoichiometric ratio between K2CrO4 and PbCrO4 is 1:1. Therefore, the moles of PbCrO4 formed will be equal to the moles of K2CrO4.
moles of PbCrO4 = 1.5 moles
Step 3: Calculate the molar mass of PbCrO4:
The molar mass of PbCrO4 can be calculated by adding the atomic masses of its constituent elements.
Molar mass of PbCrO4 = (atomic mass of Pb) + (atomic mass of Cr) + 4 x (atomic mass of O)
Molar mass of PbCrO4 = (207.2 g/mol) + (52.0 g/mol) + 4 x (16.0 g/mol)
Molar mass of PbCrO4 = 323.2 g/mol
Step 4: Calculate the mass of PbCrO4:
To calculate the mass of PbCrO4, we can use the formula:
mass = moles x molar mass
mass of PbCrO4 = 1.5 moles x 323.2 g/mol
mass of PbCrO4 = 484.8 grams
Therefore, approximately 484.8 grams of PbCrO4 will precipitate out from the reaction.
To find the number of grams of PbCrO4 that will precipitate out, you need to use stoichiometry and the given information about the solution.
To begin, let's calculate the number of moles of K2CrO4 in the solution:
Volume of solution = 500.0 milliliters = 0.5000 liters
Concentration of K2CrO4 = 3.0 M (moles per liter)
Number of moles of K2CrO4 = 0.5000 liters * 3.0 moles/liter
Now, we can use the balanced chemical equation to determine the stoichiometry between K2CrO4 and PbCrO4. From the equation:
1 mole of K2CrO4 reacts with 1 mole of PbCrO4
Therefore, the number of moles of PbCrO4 that will precipitate out is equal to the number of moles of K2CrO4 in the solution.
Now we can convert moles of PbCrO4 to grams. The molar mass of PbCrO4 is:
Molar mass of PbCrO4 = (1 * atomic mass of Pb) + (1 * atomic mass of Cr) + (4 * atomic mass of O)
Calculate the molar mass of PbCrO4 using the periodic table.
Finally, to find the mass of PbCrO4:
Mass of PbCrO4 = Number of moles of PbCrO4 * Molar mass of PbCrO4
Plug in the calculated values to get the final answer.