If you burn 33.7g of hydrogen and produce 301g of water, how much oxygen reacted?

Question

In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor.

Hydrogen + Oxygen —> Water

If you burn 33.7g of hydrogen and produce 301g of water, how much oxygen reacted?

Answer 1

Dr. Bob has shown you the detailed method of calculating the mass of O₂ that was used to produce water, however, if you recall the concept of conservation of mass, I believe you can simply subtract the mass of H₂ from the mass of water to find the mass of O₂. Thus...

301 g - 33.7 g = 267.3 g = 267 g of O2 must have been used.

Answer 2

Hello, Jada,

First balance the equation. 

2H2 + 1O2 = 2H2O

Then find the molar masses and calculate the number of moles of both H2 and H2O with the given masses.

	2	H2	1	O2	=	2	H2O
g/mole		2		32			18
Grams =		33.7		267.556			301
moles =		16.85		8.36111			16.72

Moles = grams/molar mass

·        Water at 301 grams is 16.7 moles of water.

·        Hydrogen at 33.7g is 16.9 moles of H2.

Theoretically, those two numbers should be equal since the molar ratio between H2 and H20 in the equation is equal to one (2/2). The H2O number is less than the 16.85 moles of H2, indicating some of the hydrogen did not react. Therefore, we will use the moles of water to determine O2 consumption.

The ratio of O2 to H2O in the balanced equation is ½. Therefore, ½ mole of O2 is consumed for every mole of H2O produced. That leads to 8.36 moles of O2 consumed to produce the 16.7 moles of H2O.

Convert 8.36 moles of O2 to grams O2 by multiplying by the molar mass of O2: 8.36moles*32g/mole = 268 grams.

I hope this is clear,

Bob