7. If you start with 3.20 mol of Li3N and 9.0 mol water, how many moles of NH3 will be formed?

Chemistry

1 answer:

pickupchik [31]2 years ago

7 0

Taking into account the reaction stoichiometry, 3 moles of NH₃ will be formed.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Li₃N + 3 H₂O → NH₃ + 3 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Li₃N: 1 mole
H₂O: 3 moles
NH₃: 1 mole
LiOH: 3 moles

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Li₃N reacts with 3 moles of H₂O, 3.20 moles of Li₃N reacts with how many moles of H₂O?

$moles of H_{2} O=\frac{3.2 moles of Li_{3}Nx 3 moles of H_{2} O}{1 mole of Li_{3}N}$

<u><em>moles of H₂O= 9.6 moles</em></u>

But 9.6 moles of H₂O are not available, 9 moles are available. Since you have less moles than you need to react with 3.20 moles of Li₃N, H₂O will be the limiting reagent.

<h3>Mass of NH₃ formed</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of H₂O form 1 mole of NH₃, 9 moles of H₂O form how many moles of NH₃?

$molesof NH_{3} =\frac{9 moles H_{2} Ox1 mole of NH_{3}}{3 moles H_{2} O}$