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### Topic: Percentages in chemical equation  (Read 3265 times)

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#### Geddoe ##### Percentages in chemical equation
« on: October 08, 2009, 11:05:47 AM »
ZnCl2 reacts with AgNO3 in aqueous solutions to produce a precipitate of AgCl that can be filtered and
weighed. The balanced equation for the reaction is:
ZnCl2 (aq) + 2 AgNO3 (aq) -> 2 AgCl (s) + Zn(NO3)2 (aq)
Suppose you have a mixture that contains ZnCl2, plus other compounds that do not react with AgNO3. If 0.4490
g of the mixture yields 0.6390 g of AgCl, what is the percentage of ZnCl2 in the mixture?

To be honest I'm really not sure how I should look at a question like this. I did this so far:

0.6390 g of AgCl * (35.45g Cl / 1 mole Cl) = 22.65255g Cl

I'm not sure if that is a useful step or not or what the next one shall be that I take. Any guidance is appreciated.

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#### DrCMS

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« Reply #1 on: October 08, 2009, 11:38:57 AM »
To be honest I'm really not sure how I should look at a question like this. I did this so far:

0.6390 g of AgCl * (35.45g Cl / 1 mole Cl) = 22.65255g Cl

I'm not sure if that is a useful step or not or what the next one shall be that I take. Any guidance is appreciated.

No its not a useful step and I have no idea what you are doing here at all.

Think about it again you have a known weight of unknown composition.

ZnCl2 (aq) + 2 AgNO3 (aq) -> 2 AgCl (s) + Zn(NO3)2 (aq)

From this equation you know that for every mole of ZnCl2 in the sample you'll get 2 moles of AgCl solid.

You are told the sample gives 0.6390g of AgCl.

How many moles of AgCl is that.
How many moles of ZnCl2 does that mean were in the sample?
What is the weight of the ZnCl2 calculated from the no. of moles?
That weight of ZnCl2 divided by the sample weight give the %ZnCl2 in the sample.

#### Geddoe ##### Re: Percentages in chemical equation
« Reply #2 on: October 08, 2009, 12:12:19 PM »
0.6390g of AgCl.

How many moles of AgCl is that.

0.6390g AgCl * (1 mole AgCl / 143.35) = 0.004457621 mol

How many moles of ZnCl2 does that mean were in the sample?

0.004457621 mol AgCl * (1 mole ZnCl2 / 2 mole AgCl) = 0.002228811 mole ZnCl2

What is the weight of the ZnCl2 calculated from the no. of moles?

0.002228811 * (100.84  / 1 Mole ZnCl2) = 0.224753301g ZnCl2

Sample weight calculation:

0.224753301g / 0.4490g =0.500564145

would I multiply this value by 100% ?

Is this the correct method?
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#### DrCMS

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« Reply #3 on: October 08, 2009, 01:41:23 PM »
Yes and Yes