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Topic: solution stoichimetry  (Read 2764 times)

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Offline Mr.Konami

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solution stoichimetry
« on: November 04, 2008, 12:18:49 PM »
Hello , I need your help with this question

calculate the amount of precipate formed when 22.75Ml of 0.820M silver nitrate with excess calcium chloride

2AgNO3 + cacl2---> 2Agcl + ca(No3)2

Thank you so much

For me , Lets say M: 0.820 for silver nitrate , we got the molare massa for silver nitrate then the number of moles . i am french educated so i don't know what they meant by this question


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Offline Astrokel

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Re: solution stoichimetry
« Reply #1 on: November 04, 2008, 01:53:52 PM »
hey, the question asked to calculate amount of precipitae which is the mole of precipitate. Which product looks like a precipitate to you? This is a mole ratio question.
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Offline Mr.Konami

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Re: solution stoichimetry
« Reply #2 on: November 04, 2008, 03:46:04 PM »
hello , thanks , its 2Agcl  ::)
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Offline sjb

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Re: solution stoichimetry
« Reply #3 on: November 04, 2008, 04:47:42 PM »
OK, so there are 22.75 Ml of an 0.820 M solution of silver nitrate. How many moles of silver nitrate is that?

From your balanced equation, how many moles of silver chloride does that make? So what mass of silver chloride is that?

(Not sure if you need to take into account the solubility of AgCl or not, but let's see where we go with this)
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