Question

In the reaction below, what is the limiting reactant when 1.24 moles NH3 of reacts with 1.79 moles of NO?

Answer 1

Answer:

Option 1. NO

Explanation:

The balanced equation for the reaction is given below below:

4NH₃ + 6NO —> 5N₂ + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted with 6 moles of NO.

Therefore, 1.24 moles of NH₃ will react with = (1.24 × 6)/4 = 1.86 moles of NO

From the calculation made above, we can see that a higher amount of NO (i.e 1.86 moles) than what was given (i.e 1.79 moles) is needed to react completely with 1.24 moles of NH₃.

Therefore, NO is the limiting reactant and NH₃ is the excess reactant.

Thus, the 1st option gives the correct answer to the question

Answer 2

Answer:

1. NO .

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to identify the limiting reactant by simply calculating the moles of any product, say N2, via the moles of each reactant and including the corresponding mole ratio (4:5 and 6:5):

$1.24molNH_3*\frac{5molN_2}{4molNH_3}=1.55molN_2 \\\\1.79molNO*\frac{5molN_2}{6molNO}=1.50molN_2$

Thus, since NO yields the fewest moles of N2 product, we infer it is the limiting reactant.

Regards!