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].
