Answer to Question #104786 in General Chemistry for Savannah

Question

how to find how many grams a precipitate has

Answer 1

If you do not know the exact amounts of the starting substances, you will have to act empirically. That is, first separate the sediment from the solution (by filtering or using a conventional funnel, or using a Buchner funnel). Then dry it thoroughly and weigh it on an analytical scale. This way you will get a fairly accurate result.

Well, if you know the exact amounts of substances that reacted, then everything will be much easier. For example, initially there were 28.4 grams of sodium sulfate and 20.8 grams of barium chloride. How many grams of sediment was formed?

Write the correct chemical reaction equation: Na2SO4 + BaCl2 = BaSO4 + 2NaCl.As a result of this reaction, an almost insoluble substance is formed – barium sulfate, which instantly falls out as a dense white precipitate.

Calculate which of the substances was taken in the lack, and which-in excess. To do this, calculate the molar masses of the initial reagents:46 + 32 + 64 = 142 g/mol - molar mass of sodium sulfate;

137 + 71 = 208 g/mol is the molar mass of barium chloride.That is, 0.2 mol of sodium sulfate and 0.1 mol of barium chloride entered into the reaction. The sodium sulfate were taken in excess, therefore, all barium chloride has reacted.

Calculate the amount of sediment formed. To do this, divide the molecular weight of barium sulfate by the molecular weight of barium chloride and multiply the result by the amount of the starting substance:20.8 * 233/208 = 23.3 grams.

What if sodium sulfate was in short supply? Let's assume that the reaction would not be 28.4 grams of this salt, but 5 times less – only 5.68 grams. And there is absolutely nothing complicated. 5.68 grams of sodium sulfate is 0.04 mol. Consequently, only 0.04 mol of barium chloride, i.e. 0.04 x 208 = 8.32 grams, could also react with this amount of this salt. Only 8.32 grams of the original 20.8 grams reacted.

Multiplying this value by the ratio of molar masses of barium sulfate and barium chloride, you get the answer: 8.32 * 233/208 = 9.32 grams of sediment.