If 62.7 g N2 react with 23.8 g H2 according to the reaction below, how many grams of ammonia (NH3) can be produced, and how many grams of the excess reactant will be left over?
Unbalanced equation: N2 + H2 → NH3
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Do you know how to balance the equation?
The first thing we have to do is balance this so we get N2+3H2---->2NH3
Then we can determine our limiting reagent:
mol of N2= 62.7g * (1mol/ 28g/mol)=2.24mol
mol of H2=23.8g * (1mol/2.02g/mol)=11.78 mol
Now we need to take the ratio into consideration:
1 mol of N2 makes 2 mol of NH3 so the ratio is 1:2
This means that since I have 2.24 mol of N2 I can make 4.48 mol of NH3
3 mol of H2 makes 2 mol of NH3 so the ratio is 3:2
This means that since I have 11.78mol go H2 I can make 7.85 mol of NH3
So my limiting reagent is N2 as it makes lessNH3 so that means that H2 is the excess reagent. To determine how much is left you can do the following:
2.24 mol N2 * (3mol H2 /1 molN2) * (2.02g/mol H2 / 1mol H2)=13.57g H2 was used
To find out how much was left you take 23.8-13.57= your answer
@photon336 So the balance equation would be N2+3H2 -->2NH3?
So the next thing you need to do is to figure out how many moles of each that you have.
Grams of N2 x (1mol N2/28g) = mol N2
Grams of H2 x (1 mol H2/2g) = mol H2
This will give us the number of moles that we HAVE not the number of moles we NEED to run the reaction.