Given:
Mass of (NH4)2HPO4 formed = 100g

To Find:
Mass of P2O5 initially taken

Explanation:

We are given that P2O5 is treated with H2O followed by excess NH4OH from (NH4)2HPO4. In this reaction, (NH4)2HPO4 acts as the limiting reagent and P2O5 is the reactant.

To determine the mass of P2O5 initially taken, we need to use stoichiometry and the concept of limiting reagents.

Step 1: Balanced Chemical Equation:
The balanced chemical equation for the reaction can be written as:
P2O5 + 3H2O + 2NH4OH → 2(NH4)2HPO4

Step 2: Stoichiometry:
From the balanced equation, we can see that the molar ratio between P2O5 and (NH4)2HPO4 is 1:2. This means that for every 1 mole of P2O5 used, 2 moles of (NH4)2HPO4 are formed.

Step 3: Calculation:
1. Calculate the molar mass of (NH4)2HPO4:
Molar mass of (NH4)2HPO4 = (2 * 14.01) + (4 * 1.01) + 31.00 + 16.00 + (4 * 1.01) = 132.08 g/mol

2. Calculate the number of moles of (NH4)2HPO4 formed:
Number of moles = Mass / Molar mass = 100g / 132.08 g/mol = 0.757 mol

3. Since the molar ratio between P2O5 and (NH4)2HPO4 is 1:2, the number of moles of P2O5 initially taken can be calculated as:
Number of moles of P2O5 = 0.757 mol / 2 = 0.3785 mol

4. Calculate the mass of P2O5 initially taken:
Mass of P2O5 = Number of moles of P2O5 * Molar mass of P2O5
= 0.3785 mol * (2 * 31.00 + 5 * 16.00)
= 0.3785 mol * 283.94 g/mol
= 107.53 g

Therefore, the mass of P2O5 initially taken is 107.53 grams.