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

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Discussion Overview

The discussion revolves around a stoichiometry problem involving the reaction of hydrogen with nitrogen, specifically how to calculate the mass of hydrogen needed to react with 50 grams of nitrogen. The scope includes homework-related problem-solving and mathematical reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that 16.1 grams of hydrogen are necessary for the reaction with 50 grams of nitrogen.
  • Another participant requests to see the work behind the calculations.
  • Multiple participants present calculations involving molar masses and conversion factors, with one showing a calculation that includes the molar mass of nitrogen as 28 grams due to its diatomic nature.
  • There is a correction regarding the chemical formula, with one participant noting that hydrogen should be represented as H2 instead of H3.
  • A later post questions whether the molar mass should be adjusted further, suggesting a potential mass of 32.14 grams for hydrogen.
  • Another participant confirms that the volume of one mole of any gas at standard temperature and pressure is 22.4 liters, regardless of the gas type.

Areas of Agreement / Disagreement

Participants express differing views on the correct representation of the chemical reaction and the calculations involved. There is no consensus on the final answer or the correct approach to the problem.

Contextual Notes

Some calculations presented are based on assumptions about molar masses and the diatomic nature of nitrogen. There are unresolved questions regarding the accuracy of the chemical equation and the implications for the stoichiometric calculations.

Who May Find This Useful

Students and individuals interested in chemistry, particularly those studying stoichiometry and gas laws, may find this discussion relevant.

lolecules
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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, [tex]H_2[/tex], not [tex]H_3[/tex]. 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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