Writing the net equation for a sequence of reactions

Question

1st step: CaC2(s)+2H2O(g) --> C2H2(g)+Ca(OH)2(s)

2nd step: 6C2H2(g)+3CO2(g)+4H2O(g) --> 5CH2CHCO2(g)

 

Can someone please help me figure out how to write the net chemical equation? Any tips or rules for other problems like this?

 

Thanks

Answer 1

I already answered this question in a video, but now I'm deleting that video and switching to this written answer.

Why hadn't this question been answered in the last 3.5 years? Because there's a mistake in the question that makes it impossible to answer as written. Here are the steps as given:

1. CaC₂(s) + 2 H₂O(g) ---> C₂H₂(g) + Ca(OH)₂(g)
2. 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) ---> 5 C₂H₂CHCO₂(g)

The problem is with the second step. If you check the second step for being balanced, you will find that it is balanced for C and O but not for H; there are 20 on the left but only 15 on the right.

There is only one product for the second step, but 5 molecules of it as the equation is currently written. Thus, if the molecule produced had one more H in it, the second step would be balanced for H and remain balanced for C and O without changing any coefficients.

That turns out to be the right thing to do. If one draws the structure suggested by the way the formula of the product is written, one will find that for the CO₂ at the right end of the formula, one of the Os will have only one covalent bond when it should have two, and having it bond with an H fixes that. I had tried to illustrate that for you via typing atoms and bonds, but each time I submit my answer their positions get messed up, so I've given up on that.

We already have two reasons for thinking this is the right fix, namely that it makes the second step balanced and it makes the bonding make sense. Here is one more very strong reason: the additional H turns a substance that is not known to exist into one that is, namely acrylic acid.

Now that we have that fixed, we can answer the original question. We need to combine the equations for the two steps into the overall net equation. Here's how we do that:

1. First, identify the substance that is a product of the first step and a reactant in the second step.
2. Then multiply the steps by positive integers to make the number of molecules, or formulas if it is not a molecular substance, or moles if you are thinking in those terms, of that substance the same in both steps.
3. Finding the right multipliers and resulting amounts is similar to determining least common multiples in math.
4. In applying the multipliers, remember that you are multiplying the entire step, and thus multiply all of the coefficients in the step by the applicable multiplier.
5. Add the resulting steps to get the overall net reaction. Be sure to cancel out the substance produced in the first step and used up in the second step; that is what makes the resulting equation the net equation for the reaction.

It sounds a little complicated, but actually it's usually quite easy. Here are our steps again, with the formula fixed in step 2:

1. CaC₂(s) + 2 H₂O(g) ---> C₂H₂(g) + Ca(OH)₂(g)
2. 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) ---> 5 C₂H₂CHCO₂H(g)

First we need to identify the substance that is a product of the first step and a reactant in the second step. In this case that substance is C₂H₂ (acetylene).

Then we need to identify the positive integers to multiply the steps by to make the amount of C₂H₂ the same in both steps. One molecule is produced in the first step and 6 are used up in the second step, so the multipliers are 6 for step 1 and 1, and thus no action needed, for step 2. Here are the steps again after applying the multipliers:

1. 6 CaC₂(s) + 12 H₂O(g) ---> 6 C₂H₂(g) + 6 Ca(OH)₂(g)
2. 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) ---> 5 C₂H₂CHCO₂H(g)

Now we are ready for the final step, which is to add the two steps together, cancelling out the equal amounts of C₂H₂ produced in the first step and used up in the second step:

1. 6 CaC₂(s) + 12 H₂O(g) ---> 6 C₂H₂(g) + 6 Ca(OH)₂(g)
2. 6 C₂H₂(g) + 3 CO₂(g) + 4 H₂O(g) ---> 5 C₂H₂CHCO₂H(g)

6 CaC₂(s) + 16 H₂O(g) + 3 CO₂(g) ---> 6 Ca(OH)₂(g) + 5 C₂H₂CHCO₂H(g)

Our final answer is above. Note that we combined the 12 H₂O from Step 1 and the 4 H₂O from Step 2 into 16 H₂O in our answer. It was essential that we did not do that combining until the final adding of equations so that we got the amounts right and kept the equations for the steps balanced.

I hope that this all made sense to you and that you will find it useful!