given the reaction 4NH3 + 3O2 =2N2 + 6H20
how many moles of N2 would form if 2.77 moles of NH3 reacted completely?
See your post above.
Consider the balanced equation below.
4NH3 + 3O2 mc004-1.jpg 2N2 + 6H2O
What is the mole ratio of NH3 to N2?
To determine how many moles of N2 would form if 2.77 moles of NH3 reacted completely, we need to use the balanced equation of the reaction: 4NH3 + 3O2 → 2N2 + 6H2O.
From the balanced equation, we can see that 4 moles of NH3 react to produce 2 moles of N2.
Let's set up a proportion to find the number of moles of N2 formed:
(2 moles N2 / 4 moles NH3) = (x moles N2 / 2.77 moles NH3)
Cross-multiplying and solving for x, we get:
x = (2 moles N2 × 2.77 moles NH3) / 4 moles NH3
x = 1.385 moles N2
Therefore, if 2.77 moles of NH3 reacted completely, 1.385 moles of N2 would form.
To determine the number of moles of N2 that would form when 2.77 moles of NH3 react completely, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is: 4NH3 + 3O2 → 2N2 + 6H2O
From the balanced equation, we can see that 4 moles of NH3 produce 2 moles of N2.
Now, we can set up a proportion using the given information:
4 moles NH3 / 2 moles N2 = 2.77 moles NH3 / x moles N2
We can solve for x, which represents the number of moles of N2:
4/2 = 2.77/x
Cross multiplying, we get:
4x = 2.77 * 2
4x = 5.54
Dividing both sides by 4, we find:
x = 5.54 / 4
x ≈ 1.39
Therefore, approximately 1.39 moles of N2 would form if 2.77 moles of NH3 reacted completely.