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.

Answer 1

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 = ?

Answer 2

What is the density of a sample that is 4.2 mL and 53.1 g

Answer 3

2.05