In the reaction K2CrO4 (aq) + PbCl2 (aq) 2KCl (aq) + PbCrO4 (s), how many grams of PbCrO4 will precipitate out from the reaction between 25.0 milliliters of 3.0 M K2CrO4 in an excess of PbCl2?

We first calculate the number of moles of K2CrO4 by multiplying the volume with the molarity.

n = (25/1000) x (3 M) = 0.075 mol K2CrO4

Then, from the balanced chemical equation, each mole of K2CrO4 will produce 1 mole PbCrO4 with molar mass of 323.2 g/mol.

(0.075 mol K2CrO4) (1 mol PbCrO4/1 mol K2CrO4) x (323.2 g/mol)

= 24.24 grams PbCrO4

Okay what I did was taking the given 25 milliliters and calculating this into liters (1l=1000ml) since we have to calculate always in liters, so therefore 25ml:1000=0.025l.

Now you want to find the moles in K2CrO4, with 0.025 liters and 3 M given of it. M=n/V, (Molarity equals Moles divided by Volume) and when you algebraicly transform so that "n" is on the other side you got MxV=n. Inserting the given 3M (M) and 0.025l (V) we calculate 3 x 0.025l, which equals 0.075 mol (n).

Now that we have the moles we can calculate the grams of PbCrO4 precipitating out by simply taking the molar mass of PbCrO4 (look at the periodic table and add the molar masses of the elements together, which is 207.2 mol Pb + 51.996 mol Cr + 4x 15.999 mol O = 323.192 mol PbCrO4) and now we calculate the mol of K2CrO4 (0.075 mol) by the mol of PbCrO4 (323.192 mol), so

0.075mol x 323.192mol = 24.2394 grams