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What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)?

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Answer:

The products of 1.0 kg of calcium phosphate reacting with 1.0 kg concentrated sulfuric acid (98% H2SO4 by mass)  are:

  • 1.3 kg of Calcium sulfate.
  • 0.6 kg of Phosphoric acid
  • And 0.03 kg of unconsumed sulfuric acid and 0.02 kg of impurities in the sulfuric acid.

Explanation:

Step 1.

The balanced equation of the reaction is required. For this case is:

[tex]3H_{2}SO_{4} + Ca_{3}(PO_{4})_{2} → 3CaSO_{4}+2H_{3}PO_{4}[/tex]

Where:

  • [tex]H_{2}SO_{4}[/tex] is Sulfuric Acid.
  • [tex]Ca_{3}(PO_{4})_{2}[/tex] is Calcium Phosphate.
  • [tex]CaSO_{4}[/tex] is Calcium Sulfate.
  • [tex]H_{3}PO_{4}[/tex] is Phosphoric acid.

Step 2.

The molar masses (the weight of one mol) of each compound are required:

  • Sulfuric Acid: 98.079 g/mol
  • Calcium Phosphate: 310.18 g/mol
  • Calcium Sulfate: 136.14 g/mol
  • Phosphoric acid: 97.994 g/mol

Step 3.

The amount of moles of sulfuric acid and calcium phosphate avalaible are calculated using the corresponding molar mass.

  • For sulfuric acid: [tex](1000*0.98) g  *(\frac{1 mol}{98.079 g})=9.99 mol[/tex] (note that the real amount of acid is the purity times the weight)
  • For calcium phosphate: [tex]1000 g *(\frac{1 mol}{310.18 g}) =3.22 mol[/tex].

Step 4.

From the balanced equation in Step 1. It is known that:

3 moles of Sulfuric acid need 1 mol (no coefficient) of Calcium phosphate to react.

Then is possible to determinate which of the reagents is first totally consumed (limiting reagent).

  • For sulfuric acid: For 3 moles of sulfuric acid only one mol of calcium phosphate is required. Then the 9.99 moles would need 3.33 moles of calcium phosphate (more than available).
  • For calcium phosphate: 1 mol of calcium phosphate will require 3 moles of sulfuric acid. Then, the 3.22 moles would need 9.67 moles of sulfuric acid (less than the 9.99 available).

Then, is possible to conclude that all the calcium phosphate would be consumed (It is the limiting reagent), and some sulfuric acid would remain.

Step 4.

Using the determined limiting reagent and, from the balanced equation of step 1, it is possible to know that:

For each 1 mol of calcium phosphate consumed: 3 moles of calcium sulfate and, 2 moles of phosphoric acid would be produced.

With this information:

  • Calcium sulfate: [tex]3.22*(\frac{3mol}{1mol})=9.66 mol[/tex] produced.
  • Phosphoric acid: [tex]3.22 mol*(\frac{2mol}{1mol})=6.44 mol[/tex] produced.
  • Sulfuric acid: As there is more sulfuric acid (9.99 mol) that the 9.67 required, then 0.32 mol remains after the calcium phosphate is depleted.

Step 5.

Using the molar masses in the step 2, is possible to calculate the weight of each species in the products.

  • Calcium sulfate: [tex]9.66mol*(\frac{136.14g}{1mol})=1.315g=1.3 kg[/tex] produced.
  • Phosphoric acid: [tex]6.44 mol*(\frac{97.994g}{1mol} )=631.08 g=0.6kg[/tex] produced.
  • Sulfuric acid: [tex]0.32 mol*(\frac{98.079g}{1 mol} )=31.38g=0.03kg[/tex] unused and 0.02 kg of impurities in the original sulfuric acid (98% acid, and 2% impurities).

The masses of calcium sulfate and phosphoric acid produced are 1,316.75 grams and 631.86 grams respectively.

Given:

1.0 kg calcium phosphate with 1.0 kg concentrated sulfuric acid (98% [tex]H_2SO_4[/tex] by mass)

To find:

The masses of calcium sulfate and phosphoric acid can be produced

Solution:

Mass of calcium phosphate = 1.0 kg = 1000 g

Moles of calcium phosphate :

= [tex]\frac{1000 g}{ 310.18 g/mol}=3.224 mol[/tex]

Mass of concentrated sulfuric acid solution = 1.0 kg = 1000 g

Mass percenatge of sulfuric acid solution =98%

Mass of sulfuric acid :

= [tex]1000 g\times \frac{98}{100}=980 g[/tex]

Moles of sulfuric acid:

= [tex]\frac{980 g}{98.079 g/mol}=9.992 mol[/tex]

[tex]Ca_3(PO_4)_2+3H_2SO_4\rightarrow 3CaSO_4+2H_3PO_4[/tex]

According to reaction, 1 mole of calcium  phosphate reacts with 3 moles of sulfuric acid, then 3.224 moles of calcium phosphate will react with:

[tex]=\frac{3}{1}\times 3.224 mol=9.672\text{ mol of} H_2SO_4[/tex]

As per the question, we have 9.992 moles of sulfuric acid which is more than 9.672 moles. This indicates that:

  • Sulfuric acid is an excessive reagent.
  • Calcium phosphate is a limiting reagent.

The mass of calcium sulfate and phosphoric acid will depend upon the moles of calcium phosphate.

[tex]Ca_3(PO_4)_2+3H_2SO_4\rightarrow 3CaSO_4+2H_3PO_4[/tex]

According to reaction, 1 mole of calcium phosphate gives with 3 moles of calcium sulfate, then 3.224 moles of calcium phosphate will give:

[tex]=\frac{3}{1}\times 3.224 mol=9.672\text{ mol of} CaSO_4[/tex]

Mass of 9.672 moles of calcium sulfate ;

[tex]=9.672 mol\times 136.14 g/mol=1,316.75 g[/tex]

According to reaction, 1 mole of calcium phosphate gives with 2 moles of phosphoric acid, then 3.224 moles of phosphoric acid will give:

[tex]=\frac{2}{1}\times 3.224 mol=6.448\text{ mol of} H_3PO_4[/tex]

Mass of 6.448moles of phosphoric acid;

[tex]=6.448mol\times 97.994 g/mol=631.86 g[/tex]

The masses of calcium sulfate and phosphoric acid produced are 1,316.75 grams and 631.86 grams respectively.

Learn more about limiting reagent and excessive ragent here:

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