Often happens so that during chemical reaction the slightly soluble substance which is dropping out in a deposit is formed (for example, barium sulfate, calcium phosphate, silver chloride, etc.). Let's assume, to the chemist the task is set: to determine the mass of this deposit. How it is possible to make it?
Instruction
1. If exact amounts of initial substances are unknown to you, then it is necessary to work by practical consideration. That is at first separate a deposit from solution (by filtering or on a usual funnel, or with use of a funnel of Byukhner). Then carefully dry up it and weigh on analytical scales. So you receive rather exact result.
2. And if exact amounts of the substances reacting are known to you, then everything will be much simpler. For example, initially there were 28.4 grams of sulfate of sodium and 20.8 grams of chloride of barium. How many grams of a deposit were formed?
3. Write the correct equation of chemical reaction: Na2SO4 + BaCl2 = BaSO4 + 2NaCl. As a result of this reaction almost insoluble substance – the barium sulfate which is instantly dropping out in the form of a dense white deposit is formed.
4. Calculate what of substances was taken in a shortcoming and what – is a lot of. For this purpose count the molar mass of initial reagents: 46 + 32 + 64 = 142 g/mol – the molar mass of sulfate of sodium; 137 + 71 = 208 g/mol – the molar mass of chloride of barium. That is reacted 0.2 asking sulfate of sodium and 0.1 asking barium chloride. Sulfate of sodium was taken much, therefore, all chloride of barium reacted.
5. Count quantity of the formed deposit. For this purpose divide the molecular mass of sulfate of barium into the molecular mass of chloride of barium and increase result by amount of initial substance: 20.8 * 233/208 = 23.3 grams.
6. And if sulfate of sodium was in a shortcoming? Let's assume, would react not 28.4 grams of this salt, and is 5 times less – only 5.68 grams. And there is nothing difficult at all. 5.68 grams of sulfate of sodium make 0.04 asking. Therefore, with such amount of this salt could react also only 0.04 asking barium chloride, that is 0.04 x 208 = 8.32 grams. Only 8.32 grams from initial 20.8 grams reacted.
7. Having increased this size by a ratio of molar mass of sulfate of barium and chloride of barium, receive the answer: 8.32 * 233/208 = 9.32 grams of a deposit.