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# Ideal gas equation help watch

• Boron trichloride (BCl3) can be prepared as shown by the following equation.
B2O3(s) + 3C(s) + 3Cl2(g) 2BCl3(g) + 3CO(g)
A sample of boron oxide (B2O3) was reacted completely with carbon and chlorine. The two gases produced occupied a total volume of 5000 cm3 at a pressure of 100 kPaand a temperature of 298 K.
Calculate the mass of boron oxide that reacted. Give your answer to 3 significant figures. (The gas constant R = 8.31 J K–1 mol–1)

I have tried to work this out and I looked on the mark scheme which is :

P = 100 000 (Pa) and V = 5.00 x 10-3 (m3)

n = PV/RT = 100 000 x 5.00 x 10-3/8.31 x 298 = 0.202 moles (of gas produced)

Therefore 0.202/5 = 0.0404 moles B2O3 Mass of B2O3 = 0.0404 x 69.6= 2.81 (g)

I understand all of this except i don't understand why you divide 0.202 by 5. Why do i divide by 5?
1. You've worked out the moles of gaseous product, by n=PV/RT.

The stoichiometry is:

B2O3(s) + 3C(s) + 3Cl2(g) ----> 2BCl3(g) + 3CO(g)

So; you've worked out the total moles of gas, which is 2 moles of BCl3 and 3 moles of CO. 5 moles total.

Then you look at the B2O3 : Product molar ratio; 1 mole of B2O3 to 5 moles of gas, so you need to divide by 5. 'Kay?
2. because the number of moles of the products altogether is 5 and your are only finding the moles of the B2O3 which is 1. so to get the number of moles of the B2O3 you have to divide the number of moles of both of the gases (which was 0.202mol) by 5 and times by 1.
3. (Original post by Infraspecies)
You've worked out the moles of gaseous product, by n=PV/RT.

The stoichiometry is:

B2O3(s) + 3C(s) + 3Cl2(g) ----> 2BCl3(g) + 3CO(g)

So; you've worked out the total moles of gas, which is 2 moles of BCl3 and 3 moles of CO. 5 moles total.

Then you look at the B2O3 : Product molar ratio; 1 mole of B2O3 to 5 moles of gas, so you need to divide by 5. 'Kay?
Thank you!
4. (Original post by rkoria)
because the number of moles of the products altogether is 5 and your are only finding the moles of the B2O3 which is 1. so to get the number of moles of the B2O3 you have to divide the number of moles of both of the gases (which was 0.202mol) by 5 and times by 1.
Thank you!

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Updated: May 14, 2015
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