Limiting reactants step 4

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amesbruin1L
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Joined: Fri Sep 28, 2018 12:16 am

Limiting reactants step 4

Postby amesbruin1L » Wed Oct 03, 2018 5:07 pm

In lecture today, Lavelle went over limiting reactants for the example of calcium carbide reacting w/ water to form calcium hydroxide and ethyne. For step 4 for calculating the molar mass after the equation has been balance, why aren't coefficients included in finding the molar mass of the reactants or products? Is it because we want the individual molar mass of each element? sorry if this is a silly question!

Mahir_Hasan2C
Posts: 60
Joined: Fri Sep 28, 2018 12:25 am

Re: Limiting reactants step 4

Postby Mahir_Hasan2C » Wed Oct 03, 2018 5:19 pm

When finding the molar mass of a compound you don't need the coefficient because you add the masses of all atoms in the compound. If you are talking about finding grams or the mass, you would then need the coefficient for the mole to mole ratio.

Cody Do 2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: Limiting reactants step 4

Postby Cody Do 2F » Wed Oct 03, 2018 5:35 pm

Coefficients aren't included in finding the molar mass of the reactants or products because if the coefficients were included, then the molar mass of the molecule would be incorrect! Molar mass refers to the mass of ONE molecule/atom—including the coefficient in calculations would be like finding the total mass of X amounts of the molecule/atom.

The Balanced Equation: CaC2 + 2H20 --> Ca(OH)2 + C2H2 ; It's given that there is 100g of H20 and 100g of CaC2

1) Divide 100g CaC2 by the molar mass of CaC2. The molar mass of CaC2 is 64.10 g/mol (found by adding Ca's mass of 40.08 to 2*12.01, where the 2 comes from the presence of 2 carbon atoms and 12.01 is carbon's mass). You should get 1.56 moles of CaC2.

2) Divide 100g H20 by the molar mass of H20. The molar mass of H20 is 18.02 g/mol (found by adding O's mass of 16.00 and 2 hydrogen masses of 1.01 each). You should get 5.55 moles of H20.

3) Thus, we have 1.56 moles of CaC2 and 5.55 moles of H20. Now we can see that CaC2 is the limiting reactant because there is a sufficient amount of moles of H20 to do the reaction twice (Seen from the 2 moles of H20 in the balanced equation), while there is only enough moles of CaC2 for one round of the process.

4) Now is where we use the coefficient—it's used purely to convert moles of one substance into moles of another. You only need to do this if the question asks how much product can be made with a certain amount of reactants. In this case, let's find the amount of ethyne that can be produced with 100g of CaC2. Start with the 1.56 moles of CaC2 found in step 2 above and multiply it by (1 mole C2H2/1 mole CaC2). The values are pulled from the balanced equation. By doing this, the 1.56 moles of CaC2 are converted into 1.56 moles of C2H2. To find the amount of C2H2 produced, you must not convert from moles to grams using C2H2's molar mass of 26.04 (2 Carbons of mass 12.01 each, 2 hydrogens of 1.01 each). Thus, you should end up with 40.6g of ethyne produced.

Hope that answers your question!


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