Step two in limiting reactants

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605357751
Posts: 31
Joined: Wed Sep 18, 2019 12:20 am

Step two in limiting reactants

Postby 605357751 » Sun Oct 06, 2019 3:35 pm

Example 1 of section M in the fundamentals part of the textbook, the balanced equation of the reaction is 2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O.
In the step 2 the problem says to use the mole ratio derived from the stoichiometric coefficients in the balanced equation, to convert from the amount of C8H 18 molecules into the amount in moles of CO2 molecules. My question is why do we need to convert C8118 to CO2 moles rather than H20 or O2. For future problems how do we new which mole ratio to use.

Chem_Mod
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Re: Step two in limiting reactants

Postby Chem_Mod » Sun Oct 06, 2019 5:17 pm

The question asks you to find the percent yield of carbon dioxide. Therefore, you need to calculate the theoretical yield of carbon dioxide by converting the moles of octane to grams of carbon dioxide using the mole ratio of octane to carbon dioxide in the balanced chemical equation. Had the question asked you to find the theoretical yield of water or how much oxygen was consumed, then you would use their respective mole ratios to octane.

Haley Pham 4I
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Re: Step two in limiting reactants

Postby Haley Pham 4I » Sun Oct 06, 2019 5:19 pm

The question in the problem was "In one test the engine burns 1.00L of octane (of mass 702g) and produces 1.84kg of carbon dioxide. What is the percentage yield of carbon dioxide?" The problem asked us to calculate the percentage yield of carbon dioxide. It only gave us the actual yield of CO2 molecules, so we would have to convert the amount of (C8H18) molecules into the amount in moles of CO2 molecules to calculate the theoretical yield, which we would then use to obtain the percentage yield.

605357751
Posts: 31
Joined: Wed Sep 18, 2019 12:20 am

Re: Step two in limiting reactants

Postby 605357751 » Sun Oct 06, 2019 10:38 pm

okay thank you for your response!


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