chem question for chem weez

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Smooth Operater

don't bug "operatEr"!
10+ Year Member
5+ Year Member
2X + Y --> Z

IF stoichiometric quantities of X and Y are placed in a sealed, flexiable container container with an initial volume of 30 L at STP, what volume of Z will be produced? Given that XYZ are gases.

The correct answer is 10 L. But I don't understand the logic behind getting the answer. If someone can explain it, it will be great. thanks as always! djeffreyt

Senior Member
10+ Year Member
7+ Year Member
It's a bit of a tricky question since we're so used to dealing with reactions that don't go all the way to completion, but that is what is being assumed here.

basically we put 2 moles of X and 1 mole of Y into a container of 30L at STP and we assume the reaction goes to completion. Volume of a gas is dependent only on moles of gas particles present, so initially we put in 3 moles of a gas and they react to completion to form 1 mole of Z.

so if
V' = initial volume
n' = initial moles of gas
V" = final volume
n" = final moles of gas

then

V'/n' = V"/n"

30 L / 3 moles = V" / 1 mole

V" = 10 L

714guy

counting down till 5/2012
7+ Year Member
15+ Year Member
Smooth Operater said:
2X + Y --> Z

IF stoichiometric quantities of X and Y are placed in a sealed, flexiable container container with an initial volume of 30 L at STP, what volume of Z will be produced? Given that XYZ are gases.

The correct answer is 10 L. But I don't understand the logic behind getting the answer. If someone can explain it, it will be great. thanks as always! my logic . . . since we know that at STP 1 mole of gas occupies 22.414L, making 1/2 mole of gas @ STP occupying 1/2 of 22.414L. But in this case we have a total of 30L of stoichiometric quantities of X and Y, so that would mean twice as much X than Y in the 30L. so X = 20L and Y= 10L using stoichi, we can find that we'll make 10L of Z.