Ammonium phosphate fertilizer is prepared by the following reaction

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Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq)  (NH4)3PO4(s) (a)  Calculate the molar mass of ammonia, phosphoric acid and ammonium phosphate. (b)  How many moles of ammonium phosphate are present in a 5.00-pounds bag of (NH4)3PO4?
Ammonium phosphate fertilizer is prepared by the following reaction: 3NH3(g) + H3PO4(aq) ï‚® (NH4)3PO4(s)

(a)  Calculate the molar mass of ammonia, phosphoric acid and ammonium phosphate.

(b)  How many moles of ammonium phosphate are present in a 5.00-pounds bag of (NH4)3PO4?

(c)  How many grams of ammonia and phosphoric acid, respectively, are required to produce 1.00 kg of

(NH4)3PO4 if the reaction has 100% yield;

(d)  How many kilograms (kg) of ammonia and phosphoric acid, respectively, are required to produce

1.00 kg of (NH4)3PO4 if the reaction has has 85.0% yield?

The Correct Answer and Explanation is :

Let’s solve each part step by step:

(a) Calculate the molar mass of ammonia, phosphoric acid, and ammonium phosphate.

  1. Ammonia (NH₃):
    • Nitrogen (N) has an atomic mass of 14.01 g/mol.
    • Hydrogen (H) has an atomic mass of 1.008 g/mol.
    Molar mass of NH₃ = 14.01 + (3 × 1.008) = 17.034 g/mol.
  2. Phosphoric acid (H₃PO₄):
    • Hydrogen (H) has an atomic mass of 1.008 g/mol.
    • Phosphorus (P) has an atomic mass of 30.974 g/mol.
    • Oxygen (O) has an atomic mass of 16.00 g/mol.
    Molar mass of H₃PO₄ = (3 × 1.008) + 30.974 + (4 × 16.00) = 97.994 g/mol.
  3. Ammonium phosphate ((NH₄)₃PO₄):
    • From NH₄ (ammonium) and PO₄ (phosphate), we know the molar mass is calculated by adding up the masses of 3 ammonium ions and 1 phosphate ion.
    Molar mass of (NH₄)₃PO₄ = 3 × (14.01 + 4 × 1.008) + 30.974 + 4 × 16.00 = 3 × 18.034 + 30.974 + 64.00 = 149.086 g/mol.

(b) How many moles of ammonium phosphate are present in a 5.00-pound bag of (NH₄)₃PO₄?

To convert 5.00 pounds to grams: 1 lb=453.592 g1 \, \text{lb} = 453.592 \, \text{g} 5.00 lb=5.00×453.592=2267.96 g5.00 \, \text{lb} = 5.00 \times 453.592 = 2267.96 \, \text{g}

Now, use the molar mass of ammonium phosphate (149.086 g/mol) to find moles: Moles of (NH4)3PO4=2267.96 g149.086 g/mol=15.21 mol\text{Moles of } (NH₄)₃PO₄ = \frac{2267.96 \, \text{g}}{149.086 \, \text{g/mol}} = 15.21 \, \text{mol}

(c) How many grams of ammonia and phosphoric acid are required to produce 1.00 kg of (NH₄)₃PO₄ if the reaction has 100% yield?

From the balanced equation: 3 mol NH₃ : 1 mol H₃PO₄ : 1 mol (NH₄)₃PO₄3 \, \text{mol NH₃} \, : \, 1 \, \text{mol H₃PO₄} \, : \, 1 \, \text{mol (NH₄)₃PO₄}

  1. Moles of (NH₄)₃PO₄: 1.00 kg=1000 g1.00 \, \text{kg} = 1000 \, \text{g} Moles of (NH₄)₃PO₄=1000 g149.086 g/mol=6.71 mol\text{Moles of (NH₄)₃PO₄} = \frac{1000 \, \text{g}}{149.086 \, \text{g/mol}} = 6.71 \, \text{mol}
  2. Moles of ammonia (NH₃): 3 mol NH₃ : 1 mol (NH₄)₃PO₄3 \, \text{mol NH₃} \, : \, 1 \, \text{mol (NH₄)₃PO₄} Moles of NH₃=3×6.71=20.13 mol\text{Moles of NH₃} = 3 \times 6.71 = 20.13 \, \text{mol} Grams of ammonia: 20.13 mol×17.034 g/mol=342.35 g20.13 \, \text{mol} \times 17.034 \, \text{g/mol} = 342.35 \, \text{g}
  3. Moles of phosphoric acid (H₃PO₄): Moles of H₃PO₄=6.71 mol\text{Moles of H₃PO₄} = 6.71 \, \text{mol} Grams of phosphoric acid: 6.71 mol×97.994 g/mol=657.39 g6.71 \, \text{mol} \times 97.994 \, \text{g/mol} = 657.39 \, \text{g}

(d) How many kilograms (kg) of ammonia and phosphoric acid are required to produce 1.00 kg of (NH₄)₃PO₄ if the reaction has 85.0% yield?

Theoretical amounts are calculated in part (c), but now we account for the 85% yield.

  1. Ammonia required: 342.35 g÷0.85=402.94 g342.35 \, \text{g} \div 0.85 = 402.94 \, \text{g} Kilograms of ammonia=402.94 g÷1000=0.403 kg\text{Kilograms of ammonia} = 402.94 \, \text{g} \div 1000 = 0.403 \, \text{kg}
  2. Phosphoric acid required: 657.39 g÷0.85=773.99 g657.39 \, \text{g} \div 0.85 = 773.99 \, \text{g} Kilograms of phosphoric acid=773.99 g÷1000=0.774 kg\text{Kilograms of phosphoric acid} = 773.99 \, \text{g} \div 1000 = 0.774 \, \text{kg}

Summary:

  • (a) Molar masses: NH₃ = 17.034 g/mol, H₃PO₄ = 97.994 g/mol, (NH₄)₃PO₄ = 149.086 g/mol.
  • (b) Moles of ammonium phosphate in 5.00 lb: 15.21 mol.
  • (c) Grams required for 1.00 kg of (NH₄)₃PO₄ (100% yield): 342.35 g of NH₃ and 657.39 g of H₃PO₄.
  • (d) Grams required for 1.00 kg of (NH₄)₃PO₄ (85% yield): 0.403 kg of NH₃ and 0.774 kg of H₃PO₄.

I’ll now generate the image of the chemical equation for you!

Here is the image of the balanced chemical equation for the preparation of ammonium phosphate. If you need further explanations or assistance, feel free to ask!

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