Limiting Reactant Calculations in Two-Step Reactions

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VioletKo3F
Posts: 103
Joined: Sat Sep 07, 2019 12:18 am

Limiting Reactant Calculations in Two-Step Reactions

Postby VioletKo3F » Mon Sep 30, 2019 8:54 pm

I went to the workshop today but had to leave early, so if anyone can explain this problem for me, I would be really grateful :')

The following two-step reactions produces hydrogen:

Step 1: CH4 +H2O - > CO + 3H2
Step 2: CO + H2O -> CO2 +H2

Given 0.036g of methane (CH4) and 0.108g of water, how much hydrogen, in grams, is produced?

(I got 0.018g H2 but I don't know if I did it right ¯\_(ツ)_/¯)

Jessica Tran_3K
Posts: 50
Joined: Thu Jul 25, 2019 12:17 am

Re: Limiting Reactant Calculations in Two-Step Reactions

Postby Jessica Tran_3K » Mon Sep 30, 2019 9:09 pm

Hello!

First, we need to convert the given grams to mols (CH4 would be 0.0022 mols, H2O would be 0.0060 mols). If we converted both CH4 and H20 to mols of CO using the mol ratio, we can determine that CH4 is the limiting reactant since it produces less of the product CO. Now we can use the amount of CO produced in the first reaction (0.0022 mols) and multiply it by the mol ratio in the second reaction to calculate the amount of H2. Converting the mols of H2 to grams, the answer should've been 0.0045 grams of H2! I hope this helps, please let me know if you need further clarification :-)

NRobbins_1K
Posts: 54
Joined: Sat Aug 17, 2019 12:15 am

Re: Limiting Reactant Calculations in Two-Step Reactions

Postby NRobbins_1K » Mon Sep 30, 2019 9:26 pm

Since the molar ratio of CH4 to H2O is 1:1, we start by seeing how many moles of CH4 we have to begin with, and the same number of moles of water will be consumed by the first reaction. To find the moles of methane (CH4) we divide the amount in grams (.036) by the molar mass of methane (16.04) and find that we have .002244 moles of methane. Now, we find how many moles of water we started with. We divide the .108 grams of water by water's molar mass (18.015) and find that we started with .006 moles of water. Since we have more moles of water than methane and the molar ratio of the first equation is 1:1, we can see that methane is the limiting reactant in the first reaction. Thus, when the first reaction is complete and we run out of methane, we will have used .002244 moles of both methane and water. To find how much hydrogen gas we produced in the first reaction, we note that the molar ratio of the limiting reagent (CH4) to H2 is 1:3. Therefore, the moles of CH4 consumed will result in three times as many moles of H2 produced. Since .002244 moles of CH4 were consumed, .00672 moles of H2 were produced by the first reaction. To find the weight of . 00672 moles of H2 in grams, we multiply by its molar mass (2.02) and find that the weight of hydrogen gas produced is .0136 grams. Now, when we subtract the .0022 moles of water that were used up in the first reaction, we find that there are .00375 moles of water remaining that can be used for the second reaction. To find out how much carbon monoxide (CO) was produced in the first reaction, we note that since the molar ratio of CH4 to CO in the first reaction is 1:1 there were also .002244 moles of CO produced. Therefore, there are .002244 moles of CO available for the second reaction, as well as .00375 moles of water. Since the molar ratio of CO to H20 in the second reaction is also 1:1, we can tell that since there are once again more moles of water than the other reactant (CO), water is in excess and the CO is the limiting reactant. Now, to find how many grams of H2 are produced by the second reaction once all the CO is used up, we note that the molar ratio of CO to H2 is 1:1 and therefore when all .002244 moles of CO have reacted, .002244 moles of H2 will have been produced. To find the weight of .002244 moles of H2 in grams, we multiply by its molar mass (2.02) and find that the weight of hydrogen gas produced is .00453 grams. Thus, the total amount of H2 produced is .00453 grams plus .0136 grams from the first reaction for a total of .01813 grams of H2. You had the right answer, I hope this explanation helps clarify any questions!

NRobbins_1K
Posts: 54
Joined: Sat Aug 17, 2019 12:15 am

Re: Limiting Reactant Calculations in Two-Step Reactions

Postby NRobbins_1K » Mon Sep 30, 2019 9:27 pm

@Jessica_Tran you did the problem right, but you forgot to include the H2 produced by the first stage of the reaction!

William Chan 1D
Posts: 102
Joined: Sat Sep 14, 2019 12:15 am

Re: Limiting Reactant Calculations in Two-Step Reactions

Postby William Chan 1D » Mon Sep 30, 2019 9:31 pm

It appears that both chemical equations are balanced, which makes things a bit easier.

The first step is to convert both masses to moles. Given 0.036g of methane, and that the molar mass of methane is about 16g/mol, we can determine that we have about 2.25 x 10^-3 mols of methane. Given 0.108 grams of water, and that the molar mass of water is about 18g/mol, we can determine that we have about 6 x 10^-3 moles of water.

I will be solving this problem under the assumption that both equations will react to completion.

According to the first equation,

CH4 + H20 --> CO + 3H2

methane is the limiting reactant because, in this balanced chemical equation, CH4 and H20 are used up in equal ratios (1:1), and there are more mols of H20 available than CH4, 0.00600 mols and 0.00225 mols respectively. So 0.00225 mols of methane and 0.00225 mols of H20 are used up. 0.00225 mols of CO are created, and 0.00675 mols of H2 are created. This is because, in the balanced chemical equation, there is a 1:3 ratio for CH4 to H2.

This leaves about 0.00375 mols of H20 left for the second reaction, and 0.00225 mols of CO to be used for the second equation. In this reaction, H20 again is in excess, so CO is the limiting reactant. 0.00225 mols of H20 and CO will be used up, 0.00225 mols of CO2 and H2 will be created. These numbers are all the same because there is a 1:1 ratio between all the products and reactants.

If we combine the amount of hydrogen gas we got from the first reaction (0.00675 mols) to the amount we got from the second reaction (0.00225 mols), we get around 0.009 mols of hydrogen gas. The molar mass of hydrogen gas is about 2g/mol. So if we multiply the moles by the molar mass, we get around 0.018 g of hydrogen gas.

Hope this helps!


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