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Offline jspake

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Using mole ratio.. please help
« on: May 03, 2012, 10:27:32 am » Hello.. I am confused on which mole ratio to use in the following:3.0 g of magnesium was added to 12.0 g of ethanoic acid.Mg + 2CH3COOH ? (CH3COO)2Mg + H2The mass of one mole of Mg is 24 g.The mass of one mole of CH3COOH is 60 g.How many moles of hydrogen were formed?Which mole ratio should I use.. Mg : H2 or CH3COOH : H2And please tell me the reason why..Your assistance will be appreciated Logged

Offline Arkcon

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Re: Using mole ratio.. please help
« Reply #1 on: May 03, 2012, 10:40:28 am » Start by converting grams of reactants into moles. Use the chemical equation, the numbers tell you moles Mg reacts with moles of acid to produce moles of hydrogen. Question for you: do you have more of one reactant than another? You can't get more hydrogen from one or the other, you only get as much hydrogen as you would get from both. Logged Hey, I'm not judging. I just like to shoot straight. I'm a man of science.

Offline jspake

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Re: Using mole ratio.. please help
« Reply #2 on: May 03, 2012, 12:26:03 pm » Thank you for your reply..Actually I've calculated that Magnesium is in excess.So, you mean to say that if I use both the ratios, I'll get the same number of moles?Please do me favour by showing the working..Waiting for your reply Logged

Offline AWK

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Re: Using mole ratio.. please help
« Reply #3 on: May 03, 2012, 04:02:26 pm » Actually I've calculated that Magnesium is in excessThen acetic acid is a limiting reagent. Logged AWK

Offline Arkcon

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Re: Using mole ratio.. please help
« Reply #4 on: May 04, 2012, 08:28:21 am » Quote from: jspake on May 03, 2012, 12:26:03 pm
Thank you for your reply..Actually I've calculated that Magnesium is in excess.So, you mean to say that if I use both the ratios, I'll get the same number of moles?Please do me favour by showing the working..Waiting for your reply
No, you'll get the wrong answer if you use the magnesium, in this case. If its left over, how will you get hydrogen from it? And no, we don't dump answers on this forum. You will have to solve it yourself, and be ready for an examination.Now, you have x grams of CH3COOH, how many moles is that? Logged Hey, I'm not judging. I just like to shoot straight. I'm a man of science.

Offline jspake

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Re: Using mole ratio.. please help
« Reply #5 on: May 04, 2012, 12:40:21 pm » Hmm ok.. this is how I solved it:No. of moles of CH3COOH = 12/60 = 0.2mole ratio = 2 : 1 = 0.2 : XNo. of moles of H2 = 0.1Hope solved it the right way :) Logged

Offline Arkcon

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Re: Using mole ratio.. please help
« Reply #6 on: May 04, 2012, 04:48:16 pm » Good work. I hope you see where you got each coefficient. Practice some more, make up some numbers. Or pretend you have a great excess of acid and use the mole ratio for magnesium. That's probably the way they'll put this same question on your exam. So you'll have seen something like it before, but not exactly. Logged Hey, I'm not judging. I just like to shoot straight. I'm a man of science.
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