m11

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Ekin 1C
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Joined: Fri Sep 29, 2017 7:04 am

m11

Postby Ekin 1C » Tue Oct 03, 2017 9:55 pm

Can someone explain how the oxygen is the limiting reactant

Ivy Lu 1C
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

Re: m11

Postby Ivy Lu 1C » Tue Oct 03, 2017 10:13 pm

There are two equations to the problem. For the first equation, if you use all the P4 and calculate the amount of O2 needed, you get 4.47g O2. Since you originally had 5.77g O2, you still have 1.30g O2 left. For the second equation, you use 5.77g P4 again but in the context of the second equation to calculate haw much O2 is needed. When you do this, you find that you need 2.98g O2 to complete the second chemical equation. Since 2.98g O2 is more than the 1.30 g O2 that you have left, O2 is the limiting reactant. Hopefully this helps you understand why O2 is the limiting reactant.

Andrea Grigsby 1I
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Joined: Fri Sep 29, 2017 7:03 am

Re: m11

Postby Andrea Grigsby 1I » Tue Oct 03, 2017 10:21 pm

I'm assuming this is the white phosphorus to phosphorus oxide question.

P4 + 3O2 --> P4O6
masses for both the reactants = 5.77g
get the molar mass of the elements from a periodic table and divide the mass with it to get the number of moles of each particle

n (P4) = 5.77/(30.97*4)
= 0.0466
n (O2) = 5.77/(16*2)
=1.80

however, the mole ratio of these two reactants are 1:3 so n of O2 has to be divided by 3
1.80/3 = 0.0601

... by these calculations, white phosphorus would be the limiting reagent, not oxygen

Andrea Grigsby 1I
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

Re: m11

Postby Andrea Grigsby 1I » Tue Oct 03, 2017 10:32 pm

my bad i only did half the question, but yes what @Ivy Lu 1C said

Hyein Cha 2I
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Re: m11

Postby Hyein Cha 2I » Thu Oct 05, 2017 5:35 pm

Ivy Lu 1C wrote:There are two equations to the problem. For the first equation, if you use all the P4 and calculate the amount of O2 needed, you get 4.47g O2. Since you originally had 5.77g O2, you still have 1.30g O2 left. For the second equation, you use 5.77g P4 again but in the context of the second equation to calculate haw much O2 is needed. When you do this, you find that you need 2.98g O2 to complete the second chemical equation. Since 2.98g O2 is more than the 1.30 g O2 that you have left, O2 is the limiting reactant. Hopefully this helps you understand why O2 is the limiting reactant.


The part I don't understand is how can we use 5.77g of P4 again for the second equation when it is already used up in the first equation?

Ivy Lu 1C
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am

Re: m11

Postby Ivy Lu 1C » Thu Oct 05, 2017 6:25 pm

When you use 5.77g P4 again in the second question, it's just to convert it to P4O6 because you want to find how much P4O6 came out of the first equation to use in the second equation. You would use the mole ratio so that 1 mole P4 equals 1 mole of P4O6, and then solve for how much oxygen you need, which is more than the remaining amount you have after the first equation. That's why oxygen is the limiting reactant.

Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
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Re: m11

Postby Hyein Cha 2I » Thu Oct 05, 2017 7:04 pm

Ivy Lu 1C wrote:When you use 5.77g P4 again in the second question, it's just to convert it to P4O6 because you want to find how much P4O6 came out of the first equation to use in the second equation. You would use the mole ratio so that 1 mole P4 equals 1 mole of P4O6, and then solve for how much oxygen you need, which is more than the remaining amount you have after the first equation. That's why oxygen is the limiting reactant.


oh, that makes much more sense! So you're saying that using 5.77g P4 for the second equation will give the same result as first converting 5.77g P4 to moles of P4O6 AND THEN using that moles of P4O6 for the second equation right?

Ekin 1C
Posts: 21
Joined: Fri Sep 29, 2017 7:04 am

Re: m11

Postby Ekin 1C » Sun Oct 08, 2017 3:21 pm

thanks so much!


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