How many grams of hydrogen gas will be produced

If 34.0 grams of H2 reacted with 23.8 grams of O2 in order to make water, how many grams of hydrogen gas will be left over? Please help! I dont understand how to do this. Please show work.

2H2 + O2 ---> 2H2O

I think I just answered the first part of this question.

Since 0.744 moles of the oxygen react, 1.488 moles of the hydrogen must react (2:1 ratio). The mass of 1.488 moles would be:

1.488 x 2.02 = 3.00g

Therefore there would be: 34.0-3.0 = 31.0g of hydrogen remaining

You are not going to need all the hydrogen gas for the reaction.

The relative molar mass for 2H2 is 4. (H = 1)

The relative molar mass for O2 is 16. (O=8)

This shows every 4 grams of the hydrogen reacts with every 16 grams of the oxygen. It means we need four times as much oxygen as hydrogen.

Therefore, in this case only (23.8/4)= 5.95 grams of the H2 would be required

The surplus is worked out by 34 - 5.95, which is 28.05 grams.

Step1. You got that part right A balanced equation.


2H2 + O2 ---> 2H2O

Step 2. Convert all masses into moles.

moles of H2 = 32 g / 2.02 g/mol = 15.8 mol

moles of O2 = 23.8 g / 32 g/mol = 0.74375 moles

Step 3. Look at the mole ratios between H2 and O2. The equation says that for every 2 moles of H2 you only need 1/2 that many moles of O2. Well you have 15.8 moles of H2 and only 0.74375 mol of O2. Clearly you won't be burning up all the H2. Now look at it from the O2's point of view. For ever mole of O2 you will need twice that many moles of H2. So if you have 0.74375 mol of O2 you will need to burn up 2 x that or 1.4875 mol of O2. So that is what is going to happen.

Step 4. How many moles of H2 will be left over? Well just go 15.8 - 1.4875 = 14.3125 mol H2 left over.

Step 5. Now all you have to do is convert that many moles of H2 back into mass.

mass = moles x Molar Mass

14.3125 mol x 2.02 g/mol = 28.91125 grams. I'll leave it up to you to determine sig figs.