Stoichiometry Problem: Finding Mass of Hydrogen Reacting with 50g Nitrogen

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    Stoichiometry
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To determine the mass of hydrogen needed to react with 50.0g of nitrogen, the balanced chemical equation N2 + 3H2 → 2NH3 is used. The calculation shows that 16.1g of hydrogen is required, derived from the molar masses of nitrogen and hydrogen. There is a discussion about the diatomic nature of nitrogen, confirming that its molar mass is 28g/mol, not 14g/mol. A mistake in the reaction equation was noted, as hydrogen should be represented as H2 instead of H3. The volume of one mole of any gas at standard temperature and pressure is correctly stated as 22.4L, regardless of the gas type.
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Homework Statement



"How many grams of hydrogen are necessary to react completely with 50.0g of nitrogen in the below reaction?"

Homework Equations



N2+3H3 ---> 2NH3

The Attempt at a Solution



16.1g H?
 
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Show your work.
 
50gN/1 1molN/28gN 3molH/1mol N 3gH/1molH
 
lolecules said:
50gN/1 1molN/28gN 3molH/1mol N 3gH/1molH
Close enough, I didn't round off like you did.
 
I guess my only question is since N2 is diatomic, then the molar mass of the nitrogen goes from 14 to 28?
 
lolecules said:
I guess my only question is since N2 is diatomic, then the molar mass of the nitrogen goes from 14 to 28?
That's right.
 
3gH/1molH
Um, so would that also mean here the molar mass of should be changed to 6...so the answer would be 32.14g?
 
Dangit, after looking at your problem. I didn't even notice you messed up the reaction equation. It's Hydrogen gas, H_2, not H_3. Fix that and it's solved.
 
Roco, are you slipping? :)
 
  • #10
Sorry for bringing up a dead topic, but I didnt want to just start a new one for basically the same thing...I just wanted to ask...

5 L N2/1 1molN2/22.4L N2

That's how I began setting up a problem, but is 22.4 right? Should it be something different since Nitrogen is diatomic?
 
  • #11
The volume of one mol of any gas at standard temp/pressure is 22.4 I don't care if it's gaseous Uranium.

22.4
 
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