Ammonium phosphate [tex]$\left(\left( NH_4\right)_3 PO_4\right)$[/tex] is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid [tex]$\left( H_3 PO_4\right)$[/tex] with liquid ammonia.

Calculate the moles of ammonium phosphate produced by the reaction of 0.095 mol of phosphoric acid.

Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.



Answer :

To determine the moles of ammonium phosphate ([tex]$\left(\text{NH}_4\right)_3\text{PO}_4$[/tex]) produced from the reaction of 0.095 moles of phosphoric acid ([tex]$\text{H}_3\text{PO}_4$[/tex]) with ammonia ([tex]$\text{NH}_3$[/tex]), we need to follow these steps:

1. Write the balanced chemical equation for the reaction to understand the mole ratio between reactants and products:
[tex]\[ \text{H}_3\text{PO}_4 + 3 \text{NH}_3 \rightarrow \left(\text{NH}_4\right)_3\text{PO}_4 \][/tex]

From this equation, it is clear that 1 mole of [tex]$\text{H}_3\text{PO}_4$[/tex] reacts with 3 moles of [tex]$\text{NH}_3$[/tex] to produce 1 mole of [tex]$\left(\text{NH}_4\right)_3\text{PO}_4$[/tex].

2. Determine the mole ratio between phosphoric acid ([tex]$\text{H}_3\text{PO}_4$[/tex]) and ammonium phosphate (([tex]$\left(\text{NH}_4\right)_3\text{PO}_4$[/tex]):
[tex]\[ \text{H}_3\text{PO}_4 : \left(\text{NH}_4\right)_3\text{PO}_4 = 1 : 1 \][/tex]

This means that for every mole of phosphoric acid, one mole of ammonium phosphate is produced.

3. Calculate the moles of ammonium phosphate produced:
Since the mole ratio is 1:1, the moles of ammonium phosphate produced will be equal to the moles of phosphoric acid used.

[tex]\[ \text{Moles of } \left(\text{NH}_4\right)_3\text{PO}_4 = \text{Moles of } \text{H}_3\text{PO}_4 = 0.095 \text{ mol} \][/tex]

4. Convert the result to 2 significant digits:
Rounding 0.095 to 2 significant digits gives us:

[tex]\[ \text{Moles of } \left(\text{NH}_4\right)_3\text{PO}_4 \approx 0.10 \text{ mol} \][/tex]

Therefore, the moles of ammonium phosphate produced by the reaction of 0.095 mol of phosphoric acid is [tex]\(\boxed{0.10 \text{ mol}}\)[/tex].