Karen M.

asked • 07/20/20

If 128 grams of calcium carbide, CaC2, is combined with 108 g of water, is there a limiting reagent? If so which reagent is the limiting reagent?

Calcium carbide (CaC2) reacts with water to produce acetylene (C2H2):


CaC2 (s) + 2H2O (g) produces Ca (OH)2 (s) +C2H2 (g)


If 128 grams of calcium carbide, CaC2, is combined with 108 g of water, is there a limiting reagent? If so which reagent is the limiting reagent?


My answer was yes there is a limiting reagent and it is water. After doing some math I got that 256 mol of H2O would be needed. But I’m not sure this is correct. If it isn’t, can someone please show me how to correctly do it?

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J.R. S. answered • 07/21/20

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The correct way to do this is to find moles of calcium carbide present and moles of water present, and compare the two values using the coefficients in the balanced equation.

For calcium carbide we have 128 g x 1 mol/64.1 g = 1.997 moles

For H2O we have 1.8 g x 1 mol/18 g = 0.1 moles

Since it takes 2 moles H2O per mole CaC2, water will be in limiting supply


Even if you meant to write 18 g H2O, that would be 1 mole of H2O and it would still be limiting.

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Karen M.

Thank you :) however I am still confused :( this is what I did: I multiplied 128 mol CaC2 by 2 mol H2O/1 mol CaC2 which gave me 256 mol H2O. I am very lost with the whole limiting reagents stuff because my textbook explains it one way and then other videos explain it another way and I don’t know what to do :(
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07/21/20

J.R. S.

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It's 128 GRAMS not MOLES of CaC2. The 128 grams equals 1.997 moles of CaC2. Then you are correct about multiplying this value by 2 to get the moles of H2O required (2:1 mol ratio of H2O: CaC2). This comes out to ~4 moles H2O needed, but we have only 0.1 moles H2O, so it is limiting. An easy way to find the limiting reactant in any reaction is to divide the moles of each reactant by its coefficient in the balance equation and whichever comes out lower will be limiting. If that doesn't make sense, let me know and I'll provide an example. Good luck.
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07/22/20

Karen M.

I am sooo sorry but I just noticed that I had originally accidentally written 1-8 g of H2O in my question but I meant to say 108 g of H2O. Could this possibly lead to the result I had or would I still have to take the steps you have given me?
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07/22/20

J.R. S.

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108 g H2O x 1 mol/18 g = 6 mol H2O. Now, it is no longer limiting since you need only ~4 mol H2O to react with the CaC2. Now the CaC2 becomes limiting. Does this make sense? I still don't think this explains the result you obtained however.
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07/22/20

Karen M.

Yes, the math makes sense to me but I just don’t understand where the 4 came from? Also, thank you so much for helping me
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07/22/20

Irina K. answered • 07/20/20

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Based on chemical equation CaC2 and 2 H2O react in proportion 1:2,


yes, limiting reagent will be water, because CaC2 was in excessive amount 128g.


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Karen M.

Thank youuu so much! But quick question, was my conclusion of 256 mol of H2O being needed correct?
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07/20/20

Irina K.

They do not ask you how many mol water you need. Actually, 1.8 g water is 0.1 mol. If you want to find out how much water you need to react with 128 g CaC2, look at the equation. They should react in proportion 1:2. Because CaC2 is given in excessive amount or 2 mol, so water need 4 mol. The proportion will be 2:4. If we increase one reagent, so we need to increase water too. That we resolve, if we would not know how much water react.
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07/20/20

Karen M.

Thank you :) however I am still confused :( this is what I did: I multiplied 128 mol CaC2 by 2 mol H2O/1 mol CaC2 which gave me 256 mol H2O.
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07/21/20

Irina K.

Limited means amount which, for sure, will be react.It's never left over. Look at the equation:1 mol CaC2 and 2 mol H2O will participate in full. Or, 64 g/mol CaC2 and 36 g/mol H2O. That is like recepy when you bake a cake. You should follow this proportion, 64:36. But, in reality, it's given 128:1.8. You see, it's too little H2O and too much CaC2, which will be left over. as for H2O , you have 1.8 g x 1 mol/18 g = 0.1 moles You tried (128x2)/1=256 You want from proportion 128g/1mol=x g/2mol find gram of water needed for 128 g CaC2. But that is wrong proportion. This is right one: 128g/64 =x/36; then x=(128 x36) /64 =72 g water need for 128 g CaC2
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07/21/20

Karen M.

Ohh I made a mistake in my question. I meant to say 108 g of H2O. Would this change the result?
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07/22/20

Irina K.

In that case, CaC2 will be limiting, and 108 g water is in excessive amount.
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07/22/20

Irina K.

Limited means amount which, for sure, will be react.It's never left over. Look at the equation:1 mol CaC2 and 2 mol H2O will participate in full. Or, 64 g/mol CaC2 and 36 g/mol H2O. That is like recepy when you bake a cake. You should follow this proportion, 64:36. But, in reality, it's given 128:108 You see, it's too much H2O which will be left over.
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07/22/20

Karen M.

Ahh okay :)
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07/22/20

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