# Stoichiometry problem

## Homework Statement

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

N2+3H3 ---> 2NH3

## The Attempt at a Solution

16.1g H?

50gN/1 1molN/28gN 3molH/1mol N 3gH/1molH

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?

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? :)

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?

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