[monique2302]

Hydrogen sulphide gas is combusted with oxygen gas to produce sulphur dioxide gas and water vapour. If there is 74.7 L of oxygen available, what volume of hydrogen sulphide gas may be combusted if the pressure is held constant at 198 kPa and the temperature is held constant at 310 degrees celsius.

So I have my answer down below but I'm not sure if it's correct because i get the same volume for hydrogen sulphide as the volume of oxygen given.

V1 = 74.7 L
P1 = P2 = 198 kPa
V2 = ?
T1 = T2 = 583k

P1V1/T1 = P2V2/T2

(198kPa)(74.7L)(583k) / (583k)(198kPa) = 74.7L

V2 = 74.7L

[sjb]

Hydrogen sulphide gas is combusted with oxygen gas to produce sulphur dioxide gas and water vapour. If there is 74.7 L of oxygen available, what volume of hydrogen sulphide gas may be combusted if the pressure is held constant at 198 kPa and the temperature is held constant at 310 degrees celsius.

So I have my answer down below but I'm not sure if it's correct because i get the same volume for hydrogen sulphide as the volume of oxygen given.

V1 = 74.7 L
P1 = P2 = 198 kPa
V2 = ?
T1 = T2 = 583k

P1V1/T1 = P2V2/T2

(198kPa)(74.7L)(583k) / (583k)(198kPa) = 74.7L

V2 = 74.7L

What is the equation for the reaction carried out?

[RickyC]

You are correct.

[rabolisk]

You are correct.

You are incorrect.

Hydrogen sulphide gas is combusted with oxygen gas to produce sulphur dioxide gas and water vapour. If there is 74.7 L of oxygen available, what volume of hydrogen sulphide gas may be combusted if the pressure is held constant at 198 kPa and the temperature is held constant at 310 degrees celsius.

So I have my answer down below but I'm not sure if it's correct because i get the same volume for hydrogen sulphide as the volume of oxygen given.

V1 = 74.7 L
P1 = P2 = 198 kPa
V2 = ?
T1 = T2 = 583k

P1V1/T1 = P2V2/T2

(198kPa)(74.7L)(583k) / (583k)(198kPa) = 74.7L

V2 = 74.7L

P1V1/T1 = P2V2/T2 assumes that the number of moles is constant, which you cannot just assume to be the case.