Limiting Reactant Calculations  [ENDORSED]

Moderators: Chem_Mod, Chem_Admin

Allyson Charco Dis1G
Posts: 21
Joined: Fri Sep 29, 2017 7:04 am

Limiting Reactant Calculations

Postby Allyson Charco Dis1G » Fri Oct 06, 2017 10:17 am

For the following equation determine the limiting reagent if 21.4g NH3 is reacted with 42.5g of O2
First we check to see if the equation is balanced :
4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g)
Which the equation is balanced
But what would be the next step to solving this problem? How would I calculate the number of moles
Using the grams given in the problem of would we calculate the molar mass and the use stoichiometry to get moles and then divide my the smallest number. I am confused about which direction to go in.

Jean Mok 3K
Posts: 22
Joined: Sat Jul 22, 2017 3:01 am

Re: Limiting Reactant Calculations

Postby Jean Mok 3K » Fri Oct 06, 2017 11:38 am

Yeah it seems like you know the general direction to go! You would first find the molar mass of NH3 and the molar mass of O2. Then use the MM you calculated to find the number of moles of each respective reactant. Then, pick a reactant and calculate the needed amount of the other using the ratio from the balanced chemical reaction. Finally compare the needed amount and the actual amount to observe if the reactant is limiting or in excess.

Kamryn Chang 1I
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am

Re: Limiting Reactant Calculations  [ENDORSED]

Postby Kamryn Chang 1I » Fri Oct 06, 2017 6:02 pm

In this problem, you wouldn't have to divide by the smallest number of moles because you are not solving for the empirical formula :) Instead, you just need to calculate the amount of moles of O2 and NH3. You can do this by using the amount of each reactant given in the problem in grams and dividing it by the molar mass of the reactant.

For example,
molar mass of O2 = 15.9994 X 2 = 31.9988 g/mol
mol O2 = 42.5 g / 31.9988 g/mol = 1.33 mol O2

You would also calculate the amount of moles of NH3 in this problem using this method.

Jesus Rodriguez 1J
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

Re: Limiting Reactant Calculations

Postby Jesus Rodriguez 1J » Sat Oct 07, 2017 8:50 pm

Your next step would be to convert the grams of reactant into moles. You would do this by dividing the grams given to you by the molar mass of the reactant. From then you would need to divided the number moles by the coefficient in front of the particular reactant. Then whichever reactant has the least amount of moles would be your limiting reactant.


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 3 guests