If 12.85 g of chromium metal is reacted with 10.72 g of phosphoric acid, then what is the maximum mass in grams of chromium(iii) phosphate that can be produced? 2cr(s) + 2h3po4(aq) → 2crpo4(s) + 3h2(g)

16.02 g

Explanation:

We are given;

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)

Step 1: Determining the number of moles of Chromium and phosphoric acid

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

Step 2: Determine the rate limiting reactant 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.Step 3: Moles of Chromium(III) phosphateFrom 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

Step 4: Maximum mass of  CrPO₄ that can be produced

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

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