# A 100.0 mL sample of a 0.200 molar aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 molar aqueous solution of BaCl2. The mixture was stirred and the precipitate was collected, dried carefully, and weighed. How many grams of precipitate should be obtained? The reaction is shown below.

## K2CO4 + BaCl2 ==> BaCrO4 + 2KCl

mols K2CrO4 = M x L = approx = 0.02
mols BaCl2 = M x L = approx 0.01

The limiting reagent is BaCl2 so you will obtain 0.01 mols BaCrO4.
g BaCrO4 = mols x molar mass = ?

## To determine the number of grams of precipitate that should be obtained, we need to consider the reaction between K2CrO4 and BaCl2, and use stoichiometry to calculate the mass of the precipitate.

The balanced equation for the reaction is:
2K2CrO4 (aq) + 3BaCl2 (aq) → BaCrO4 (s) + 4KCl (aq)

To determine the limiting reactant, we need to determine which reactant is present in the least amount, by comparing the number of moles of each reactant.

First, we need to calculate the number of moles of each reactant:
Number of moles of K2CrO4 = concentration (molarity) × volume (in liters)
Number of moles of K2CrO4 = 0.200 mol/L × 0.100 L = 0.020 mol

Number of moles of BaCl2 = concentration (molarity) × volume (in liters)
Number of moles of BaCl2 = 0.100 mol/L × 0.100 L = 0.010 mol

Next, we need to determine the stoichiometric ratio between K2CrO4 and BaCrO4. From the balanced equation, we can see that 2 moles of K2CrO4 react with 1 mole of BaCrO4. Therefore, the ratio is 2:1.

Since the mole ratio is 2:1 and we have 0.020 moles of K2CrO4, the number of moles of BaCrO4 that can be produced is 0.020 mol ÷ 2 = 0.010 mol.

Now, we need to calculate the mass of BaCrO4:
Mass of BaCrO4 = moles of BaCrO4 × molar mass of BaCrO4

The molar mass of BaCrO4 can be calculated by adding the atomic masses of each element:
molar mass of BaCrO4 = atomic mass of Ba + atomic mass of Cr + 4 × atomic mass of O
molar mass of BaCrO4 = (137.33 g/mol) + (51.996 g/mol) + 4 × (16.00 g/mol)
molar mass of BaCrO4 = 329.33 g/mol

Finally,
Mass of BaCrO4 = 0.010 mol × 329.33 g/mol = 3.293 g

Therefore, approximately 3.293 grams of precipitate (BaCrO4) should be obtained.