<h3>
Answer:</h3>
16.02 g
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of chromium as 12.85 g
- Mass of phosphoric acid as 10.72 g
We are require to calculate the maximum mass of chromium (III) phosphate that can be produced.
- The equation for the reaction is;
2Cr(s) + 2H₃PO₄(aq) → 2CrPO₄(s) + 3H₂(g)
<h3>Step 1: Determining the number of moles of Chromium and phosphoric acid</h3>
Moles = Mass ÷ Molar mass
Molar mass of chromium = 52.0 g/mol
Moles of Chromium = 12.85 g ÷ 52.0 g/mol
= 0.247 moles
Molar mass of phosphoric acid = 97.994 g/mol
Moles of phosphoric acid = 10.72 g ÷ 97.994 g/mol
= 0.109 moles
<h3>Step 2: Determine the rate limiting reactant </h3>
- From the equation, 2 moles of chromium reacts with 2 moles of phosphoric acid.
- Therefore, 0.247 moles of Chromium will require 0.247 moles of phosphoric acid, but we only have 0.109 moles.
- This means chromium is in excess and phosphoric acid is the rate limiting reagent.
<h3>Step 3: Moles of Chromium(III) phosphate</h3>
- From the equation, 2 moles of phosphoric acid reacts to yield 2 moles of chromium (III) phosphate.
- Therefore, Moles of Chromium (III) phosphate = Moles of phosphoric acid
Hence; moles of CrPO₄ = 0.109 moles
<h3>Step 4: Maximum mass of CrPO₄ that can be produced</h3>
We know that, mass = Moles × Molar mass
Molar mass of CrPO₄ = 146.97 g/mol
Thus,
Mass of CrPO₄ = 0.109 moles × 146.97 g/mol
= 16.02 g
Thus, the maximum mass of CrPO₄ that can be produced is 16.02 g
