Please help. I have submitted this to teacher and has been graded can you help me find the right answer.?

Question

There are three that I am disputing. Here they are:

Determine the number of atoms that are in 1.25 mol of O2.

Find the percentage composition of a compound containing 32.0 g of bromine and 4.9 g of magnesium.

If your chemistry classroom has a volume of 1.8 x 10^5 dm^3, how many gas particles are in the room at standard temperature and pressure? (Remember that 1 mole of any gas occupies 22.4 dm^3 at STP.)

Thank you for your help!

This is my son's Chemistry homework. I trying to help him with it.

Answer 1

1.
(1.25 mol O2)*(6.02x10^23 molecules/mole) = 7.53x10^23 molecules O2
It would be easy to put that as the answer, but the question asks for atoms, and there are 2 oxygen atoms in 1 O2 molecule, so...
(7.53x10^23 molecules O2)*(2 atoms/molecule O2) = 1.51x10^24 Oxygen atoms
So be careful on that one - make sure the teacher also realizes he/she asked for atoms and not molecules.

2.
32 + 4.9 = 36.9g compound
(32.0 g Br2)/(36.9 g cmpd)*(100%) = 86.72% Br2
(4.9 g Mg)/(36.9 g cmpd)*(100%) = 13.28% Mg

3.
(1.8x10^5 dm^3)/(22.4 dm^3/mole)*(6.02x10^23 particles/mole) = 4.84x10^27 particles in the classroom
~~~~messed this one up the first time - but it's fixed now~~~

Answer 2

1. 1 mole = 6.022 X10^23 particles per mole (Avogadro's
Number)
1.25(6.022X 10^23)= 7.5275 X 10^23 particles.
each molecule of O2 has two atoms of oxygen, so
7.5275 X 10^23(2)= 1.5055 X 10^24 atoms of oxygen in 1.25 moles of O2.

2. 32.0 + 4.9 = 36.9 grams total.
32/36.9=0.8672 X 100= 86.72 % Bromine
4.9/36.9=0.1328 X 100= 13.28 % Magnesium

3. 1.8 X 10^5 divided by 22.4= 0.0804 X 10^5= 8.04 X 10^3 moles of gas.
8.04 X 10^3(6.022 X 10^23) = 48.4169 X 10^26
= 4.84169 X 10^27 gas particles