Question

Assuming all volume measurements are made at the same temperature and pressure, how many liters of water vapor can be produced when 8.23 liters of oxygen gas react with excess hydrogen gas? Show all of the work used to solve this problem.

Balanced equation: 2H2 (g) + O2 (g) yields 2H2 O(g)

Answer 1

@Matt6288

Answer 2

2H2 (g) + O2 (g) ---> 2H2O (g)
1 mol of O2 yields 2 mol H2O.
Likewise, 1 L of O2 yields 2 L H2O.
So, 8.23 L O2 yields 16.46 L H2O.

Answer 3

What would be the equation?

Answer 4

Are you still typing the reply?

Answer 5

Can you help me find the right answer?

Answer 6

Volume of gases are equal if there moles are equal
1 mole of O2 makes 2 moles of water vapor
So volume of water vapor would be twice the volume of O2
Volume of water vapor=8.23* 2
Volume of water vapor=16.46 liters
8.23LO2 x 2 molesH2O/1 mole O2

Answer 7

Is the answer 16.46? @Matt6288

Answer 8

@sweetburger

Answer 9

yes

Answer 10

How should i show my work??

Answer 11

@sweetburger is this right?

Answer 12

Looks good to me.

Answer 13

This is how you can show your work:
8.23LO2 x 2 molesH2O/1 mole O2=16.46L of water vapor