Wrong calculation-can' find the Lim. Reactant

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derinceltik1K
Posts: 51
Joined: Sat Aug 17, 2019 12:15 am

Wrong calculation-can' find the Lim. Reactant

Postby derinceltik1K » Sat Oct 12, 2019 2:22 pm

4NH3 + 5O2 ->4NO + 6H20

21.4 g NH3 42.5g O2

Calculate the products in grams?

I get 1.26 moles of NH3 and 1.33 mol of O2 but can't get the calculation right afterwards? How would you solve this?

Siya Shah 1J
Posts: 50
Joined: Sat Aug 17, 2019 12:15 am

Re: Wrong calculation-can' find the Lim. Reactant

Postby Siya Shah 1J » Sat Oct 12, 2019 2:32 pm

Well, considering how far you've gotten, you know that NH3 is the limiting reactant for the reaction. So you would use the mole ratio between NH3 and NO multiply by the molar mass of NO to determine how many grams of NO are produced. You would do the same for H2O.

ZainAlrawi_1J
Posts: 71
Joined: Sat Aug 24, 2019 12:16 am

Re: Wrong calculation-can' find the Lim. Reactant

Postby ZainAlrawi_1J » Sat Oct 12, 2019 2:39 pm

Since you have 1.26 moles of NH3 and 1.33 mol of O2, then you can already deduce that NH3 is the limiting reactant.
You can now use the mole ratio of 4 NH3 for every 4 NO molecules to deduce how many grams of NO are produced, and then use the same process to figure out how many grams of H2O, by using the ratio of 4 NH3 for every 6 H2O molecules.


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