During chemical reaction various substances can be formed: gaseous, soluble, slightly soluble. In the latter case they drop out in a deposit. Often there is a need to learn what exact mass of the formed deposit. How it can be calculated?
It is required to you
- - glass funnel;
- - paper filter;
- - laboratory scales.
Instruction
1. You can act by practical consideration. That is, carry out chemical reaction, carefully separate the formed deposit from a filtrate by means of a usual glass funnel and the paper filter, for example. Fuller office is reached by means of vacuum filtration (on Byukhner's funnel).
2. After that dry up a deposit – a natural way or under a vacuum, and weigh with perhaps bigger accuracy. Best of all, on sensitive laboratory scales. Here the objective will be so solved. Resort to this method when exact amounts of the initial substances which reacted are unknown.
3. If these quantities are known to you, then the problem can be solved much more simply and quicker. Let's assume, it is necessary to calculate how many chloride silver was formed in interaction of 20 grams of chloride sodium - table salt - and 17 grams of nitrate silver. First of all, write the reaction equation: NaCl + AgNO3 = NaNO3 + AgCl.
4. During this reaction very little soluble connection – the silver chloride which is dropping out in the form of a white deposit is formed.
5. Count the molar mass of initial substances. For chloride sodium it approximately makes 58.5 g/mol, for nitrate silver – 170 g/mol. That is, initially under the terms of a task you had 20/58.5 = 0.342 asking chloride sodium and 17/170 = 0.1 asking nitrate silver.
6. Thus, it turns out that chloride sodium initially was taken much, that is, reaction on the second initial substance will take place up to the end (all will react 0.1 asking nitrate silver, "having connected" the same 0.1 asking table salt). How many it is formed chloride silver? For the answer to this question, find the molecular mass of the formed deposit: 108 + 35.5 = 143.5. Having increased initial amount of nitrate silver (17 grams) by a ratio of molecular mass of a product and initial substance, receive the answer: 17 * 143.5/170 = 14.3 grams. Here such is there will be an exact mass of the deposit formed during reaction.