Chem. Gas Laws. :| Ideal Gas Eq'ns blah. :|?

Question

What weight of hydrogen at STP could be contained in a vessel that holds 4.8 g oxygen at STP?

Answer 1

4.8gO2 x 1 mole/32g O2 x 2.02g H2/1 mole = 0.303g H2

Answer 2

I think you mean, "What mass of H2 at STP will have the same volume as 4.8gO2 at STP?"

PV = nRT

and

PV = mRT/M where m is the mass and M is the molar mass

V = mRT/MP

V(H2) = V(O2)

mRT/MP(H2) = mRT/MP(O2)

m(H2) = mRT/MP(O2) / RT/MP(H2)

Cancel out R, T and P since both gases are at STP.

m(H2) = m/M(O2) / M(H2)

m(H2) = 4.8g / 32g x 2.0g = 0.30 g H2

Check:

V(H2) = mRT/M = 0.300g x 0.0821 Latm/molK x 273K / 2.0g/mol

V(H2) = 3.36 L H2

V(O2) = 4.80g x 0.0821 Latm/molK x 273K / 32.0g/mol

V(O2) = 3.36 L O2

Answer: 0.30 g H2

Answer 3

PV=nRT

n=w/M moles Since both the gases are at STP so their volume is same.

Now for oxygen ,

w=4.8 g M=32 so n= 4.8/32=.15 moles

so there will be .15 moles  of hydrogen & in weight it is= .15*2.016=.3024 g