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Dear students, in this question we have 5 beakers 1, 2, 3, 4 and 5.
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All the vickers contain same amount of silver nitrate, that is 0 .5 multiplied by 10x0 millimole which is equal to 0 .5 more.
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Now, into each of these beakers to add a certain amount of metal chloride, the amount of metal chloride are given, 0 .15mol calcium chloride, 0 .2m .2m.
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Aluminum chloride, 0 .25 more calcium chloride like this, the amounts are given.
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Now which two beakers contain the maximum mass of silver chloride? because we know when silver nitrate reacts with a metal halide or metal chloride, silver chloride precipitate will form.
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So we have to find out which two beakers contain the maximum mass of silver chloride precipitated.
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Now to solve this, let us first write the equations involved here.
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So, first one is ag no3 plus ca cl2, so agcl2 plus ca, no3, here are 2 moles, so here 2 now the other one is plus al, cl3, 3 moles of silver nitrate reacts with aluminum chloride to form al, sorry, ag, cl, so 3 moles of ag, cl, so 3 moles of ag cl, 3 moles of ag c, cl plus al no3 thrice.
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Then coming to calcium chloride again the same thing.
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So let us write the equation again to solve the problem we need this.
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Ca no3 twice.
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Then with nacl.
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With nacl, agl, agno3 plus nacl, sgl plus nano3 plus na no3.
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Now let us start with the first one.
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For everyone, we have 0 .5 mole of hg no3.
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In all the cases, hgno3 amount is the initial amount is same.
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Now, here calcium quality is 0 .15 mole.
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Now as for the equation, 1 mole of calcium chloride reacts with 2 more silver nitrate, so for 0 .15 we need 0 .3 mole.
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And because we have having 0 .5 mole, therefore calcium quad is limiting reacted.
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Now, equal to silvernated, congenital.
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Silver chloride will form.
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So, silver chloride will form is 0 .3 mole here.
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Here, aluminum chloride we have 0 .20 mole.
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0 .20 moles means as for the balanced equation, we need 0 .6 mole...