Taking into account the reaction stoichiometry, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.
Reaction stoichiometry
In first place, the balanced reaction is:
3 NaOH + H₃PO₄ → 3 H₂O + Na₃PO₄
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:
- NaOH: 3 moles
- H₃PO₄: 1 mole
- H₂O: 3 moles
- Na₃PO₄: 1 mole
Limiting reagent
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.
Limiting reagent in this case
To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of H₃PO₄ reacts with 3 moles of NaOH, 9 moles of H₃PO₄ reacts with how many moles of NaOH?
[tex]moles of NaOH=\frac{9 moles of H_{3} PO_{4} x3 moles of NaOH}{1 mole of H_{3} PO_{4}}[/tex]
moles of NaOH= 27 moles
But 27 moles of NaOH are not available, 6 moles are available. Since you have less moles than you need to react with 9 moles of H₃PO₄, NaOH will be the limiting reagent.
Moles of Na₃PO₄ formed
Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 3 moles of NaOH form 1 mole of Na₃PO₄, 6 moles of NaOH form how many moles of Na₃PO₄?
[tex]moles of Na_{3}P O_{4} =\frac{6 moles of NaOHx1 mole of Na_{3}P O_{4} }{3 moles of NaOH}[/tex]
moles of Na₃PO₄= 2 moles
Then, 2 moles of Na₃PO₄ can be produced when 6.0 mol NaOH reacts with 9.0 mol H₃PO₄.
Learn more about the reaction stoichiometry:
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