Question

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)

Answer 1

Answer:

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) phosphate

• 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

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