Fundamentals M.11 Limiting Reactants

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Marisa Gaitan 2D
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Fundamentals M.11 Limiting Reactants

Postby Marisa Gaitan 2D » Mon Oct 05, 2020 1:04 pm

Hi, I'm confused on why the limiting reactant is O2 (I thought it was P4) and am not too sure on what exactly you compare the moles to in order to figure out limiting reactants in general, thanks.

A reaction vessel contains 5.77 g of white phosphorus and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorus(III) oxide, P4 O6: P4(s)+3O2(g) → P4O6(s) If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) oxide, P4 O10: P4O6(s)+2O2(g) → P4O10(s) (a) What is the limiting reactant for the formation of P 4 O 10 ? (b) What mass of P 4 O 10 is produced? (c) How many grams of the excess reactant remain in the reaction vessel?

EnricoArambulo3H
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Re: Fundamentals M.11 Limiting Reactants

Postby EnricoArambulo3H » Mon Oct 05, 2020 1:08 pm

The limiting reactant in the first step is P4, but then you need to find out how much oxygen was leftover from the first step and use that to calculate the limiting reactant in the second step. Once you do these calculations, you will find that the limiting reactant is oxygen instead of P4O6.

John Pham 3L
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Re: Fundamentals M.11 Limiting Reactants

Postby John Pham 3L » Mon Oct 05, 2020 1:13 pm

The limiting reactant is O2 since there is a excess P4O6 remaining in the second reaction that takes place.

Initially, you calculate the amount of P4O6 that can be produced from the first reaction.
5.77 g of P4 can produce up to 0.0466 mol of P4O6
5.77 g of O2 can produce up to 0.0601 mol P4O6

Subtracting 0.0466 mol P4O6 from 0.0601 mol P4O6 would leave us with 0.0135 mol P4O6 that could be produced if there was additional P4 to react with the excess oxygen. Converting 0.0135 mol P4O6 to oxygen gives us 0.0405 mol O2 remaining.

From the calculations, you can see that white phosphorus is the limiting reactant and that oxygen is the excess reactant. However, you still aren't done with the reaction. You still have to calculate the second reaction that produces P4O10.
0.0405 mol 02 can produce 5.75 g of P4O10
0.00466 mol P4O6 can produce 13.2 g P4O10

From this, we can see that O2 is the limiting reactant in the equation and that there is excess P4O6. Basically, you need to go through both of the given equations to see how much P4O10 can be produced from the given reactants.

Hope this makes sense!

SainehaMaddineni_3I
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Re: Fundamentals M.11 Limiting Reactants

Postby SainehaMaddineni_3I » Mon Oct 05, 2020 1:17 pm

In general, to find the limiting reactant, you compare the amount of product produced by each reactant using the stoichemtric coefficients. First, you convert the reactants' masses to moles. After, based on the balanced equation, you use the mole ratio of that reactant's stoichiometric coefficient to that of the product's. You do this to find the moles of product that could be produced by that reactant. After doing this for all the reactants, you would see that your limiting reactant is the one that produced the least amount of product.

sophie esherick 3H
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Re: Fundamentals M.11 Limiting Reactants

Postby sophie esherick 3H » Mon Oct 05, 2020 1:29 pm

marisagaitan1G wrote:Hi, I'm confused on why the limiting reactant is O2 (I thought it was P4) and am not too sure on what exactly you compare the moles to in order to figure out limiting reactants in general, thanks.

A reaction vessel contains 5.77 g of white phosphorus and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorus(III) oxide, P4 O6: P4(s)+3O2(g) → P4O6(s) If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) oxide, P4 O10: P4O6(s)+2O2(g) → P4O10(s) (a) What is the limiting reactant for the formation of P 4 O 10 ? (b) What mass of P 4 O 10 is produced? (c) How many grams of the excess reactant remain in the reaction vessel?


Like other students said previously, P4 was the limiting reactant in the first step but since the question is asking you for the limiting reactants for the products of the second step, you have to use stoichiometry and dimensional analysis to get O2 as your limiting reactant for P4010. In regards for figuring out which reactant is your limiting reactant, you have to compare the moles of each reactant to the moles of the product you are told to find. To do this, you convert your reactants to moles given initial mass of your reactants and then you use the stoichiometric coefficients within your balanced chemical equation to convert from moles of one of your reactants to moles of your product. Whichever reactant yields a smaller amount of product is therefore your limiting reactant. Hope that helps!

Malakai Espinosa 3E
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Re: Fundamentals M.11 Limiting Reactants

Postby Malakai Espinosa 3E » Mon Oct 05, 2020 1:39 pm

So based on the responses, I gather the understanding that in order to find the amount of excess, we need to subtract the mols of product formed by the limiting reagent from the mols of product formed by the excess? Then we convert that number into grams of the excess reactant? Can someone clarify that I'm understanding this correctly?

SainehaMaddineni_3I
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Re: Fundamentals M.11 Limiting Reactants

Postby SainehaMaddineni_3I » Tue Oct 06, 2020 3:40 pm

Malakai Espinosa 3D wrote:So based on the responses, I gather the understanding that in order to find the amount of excess, we need to subtract the mols of product formed by the limiting reagent from the mols of product formed by the excess? Then we convert that number into grams of the excess reactant? Can someone clarify that I'm understanding this correctly?


Hi! The easier way would be using stoichiometry (the coefficients) to go from the moles of limiting reactant to moles of excess reactant. Then, convert this moles of excess reactant to grams and subtract from the given amount in the problem.

ALTERNATIVELY: To find the amount of excess reactant, you would use the moles of product produced by the limiting reactant, not the amount produced by the excess reactant. You would then use stoichiometry to see how many moles of the excess reactant would be needed to produce the same moles of product as the limiting reactant would. Convert that moles of excess reactant to grams. Then subtract it from the original amount of excess reactant to see how much would remain.

I hope this helps!


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