Limiting Reactant Practice Problems

This chemistry video tutorial provides a basic introduction of limiting reactants. It explains how to identify the limiting reactant given the mass in grams or the quantity of each reactant in moles. It explains how to use the limiting reactant to determine the moles or mass in grams of product that can be produced in a chemical reaction. This video contains plenty of examples and stoichiometry practice problems.
Stoichiometry - Free Formula Sheet:
bit.ly/3Z7uMw8
Stoichiometry - Video Lessons:
www.video-tuto...
Chemistry 1 Final Exam Review:
• General Chemistry 1 Re...
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Introduction to Moles:
• Introduction to Moles
How To Calculate the Molar Mass:
• How To Calculate The M...
How To Convert Grams to Moles:
• How To Convert Grams T...
How To Convert Moles to Grams:
• How To Convert Moles t...
Moles to Atoms Conversion:
• Moles To Atoms Convers...
Grams to Molecules Conversion:
• Grams to Molecules and...
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Grams to Atoms:
• How To Convert Grams t...
Moles, Atoms, & Grams Conversions:
• How To Convert Between...
How To Balance Chemical Equations:
• How To Balance Chemica...
Stoichiometry - Basic Introduction:
• Stoichiometry Basic In...
Avogadro's Number:
• Avogadro's Number, The...
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Limiting Reactant Problems:
• Limiting Reactant Prac...
Excess Reactant Problems:
• How To Find The Amount...
Theoretical & Percent Yield:
• How To Calculate Theor...
Percent Yield - More Examples:
• How To Calculate The P...
Percent Error:
• Percent Error Made Easy!
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Percent Composition by Mass:
• Percent Composition By...
Empirical Formula Problems:
• Empirical Formula & Mo...
Empirical Formula - Hydrated Compounds:
• How To Find The Empiri...
Combustion Analysis:
• Introduction to Combus...
Stoichiometry Practice Test:
• How To Solve Stoichiom...
_______________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections

[@TheOrganicChemistryTutor]

Stoichiometry - Free Formula Sheet: bit.ly/3Z7uMw8 Stoichiometry - Video Lessons: www.video-tutor.net/stoichiometry.html

[@christelsinoy3824]

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[@-clipz-]

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[@abhigyansrivastava4247]

Question- Do we always have to convert the grams to moles when finding limiting reagent or the theoretical yield or can we also solve the question by keeping the grams instead of converting it into moles

[@wdts7714]

Isn't Cl diatomic?

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i have a test that is litterally 50% of my entire grade in 3 hours, this helped, thanks!

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The way I screamed when I saw how easy it is to find the limiting reactant. My teachers would use very hard ways to find them and I was literally crying. (I know it's stupid lol but I was seriously struggling). You made everything so much easier. thank you!!!!

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I learned more in 5 minutes than I did with my school Chemistry teacher in 1 month!

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Chem final tomorrow and I know that this will most definitely be a major part there. I can’t figure out how to use an ICE table for limiting reagents even if my life were on the line (which, with how much my final weights, it kind of is) and this is a true life saviour for me. Thank you

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[@crazybby1543]

I get what you did to find the limiting reactant for the moles and I understand how you used the grams to find the limiting reactant, but why didn't you convert from atoms/molecules to moles to find the limiting reactant for part a?

[@NinooL807]

exactly thats what i was thinking

[@larrrr3688]

im so late lmao but it is because moles are just a set number of atoms or molecules, a mole is just an easier way of saying 6.022x10^23 atoms/molecules. There's no need to convert it because you are looking at the ratio, which is the same whether you are using atom/molecules or moles.

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first off, your videos are amazing and I can't thank you enough for your enduring clarity. I had a question about the way you balanced the equation in part a. Why can't we just multiply the oxygen molecule by 7? why must be multiply it by 7/2? I must be missing something simple...

[@a.a9021]

Not sure if you still care since it's been a while, but it's because the oxygen molecule is O2, so putting a 7 in front of it would make it 14. In order to make it match the 7 oxygen molecules on the other side, the number in front of O2 needed to be _half_ of 7 so that in the end there would be 7 oxygen molecules. Thus, 7/2. But because you can't really have fractions as coefficients, the solution was to multiply the entire equation by 2, so you'll see that in the end he actually did put a 7 in front of the oxygen molecule, and he also doubled the other coefficients.

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[@NinooL807]

easy way to find theoritical yeild the coefficent of reactant are moles now 2 moles of c2h6 is giving 4 moles of co2 1 mole would give 4/2= 2 in part a we are give how many mole of c02 will be produced if 5mole of c2h6 reacts .therefore 2*5=10mole of co2 would be produced similarly since coefficient of 02 is 7 and due to 7moles of o2 4 moles of co2 are produced now if we have 16 moles of o2 , 4/7*16= 9.1428 moles of co2 would be produced now since we know O2 is L.R Therefore we would consider 9.1428 as ans LMFAO i dont know if anyone could understand this but yeahits easier this way.

[@5kunk157h35h17]

Great video but it would be even more helpful if you explained why we the different steps are required for the second type of formula, deciding both limiting reactant and theoretical yield. As it is now, with these different steps, it becomes very abstract and difficult to remember. If the reason behind the different steps were explained, I think it would be easier to understand and remember. Anyway great video and many thanks.

[@SM-qn7sr]

Theoretical yield means the amount of product produced during the reaction. In order to find this amount it needs to depend on the limiting reactant because once the limiting reactant runs out, you can’t make anymore of the product no matter how much excess reactant you have. Going by this logic, you need to find the limiting reactant to convert to the amount of products it will produce

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[@jntwo5244]

Question: In number 1, the given quantity is in atoms for Zinc while HCl is in moles? Do you still apply the quotient of the coefficient and quantity?

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[@graydonrobson7471]

An easier and faster way to determine theoretical yield is Step 1 Determine limited reactant (given mol/coefficient)= x whichever number of smaller is your limited reactant Step 2 (Limited reactant moles given / 1) * (moles of product / Limited reactant coefficient) = theoretical yield in moles

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[@arizonahudson8536]

Why didn't you convert molecules to moles in question 1 (a) ???

[@davidg5311]

I think because its a formula unit and not grams

[@parijat8166]

@@davidg5311 you can still convert it though. I did and I got it correct. However the numbers are really small, so they are harder to deal with.

[@Christsavedm3]

Would this work every time? If you take the moles/coefficient like in 2(a). And multiply the Limiting Reactant number for O2 ~2.28 by the coefficient of CO2 which is 4 to find the moles for CO2 that would be produced about 9.14?

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