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One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 26.8 g of Ca₃(PO₄)₂ reacts with 54.3 g of H₂SO₄, what is the percent yield if 10.9 g of H₃PO₄ is formed via the UNBALANCED equation below? Ca₃(PO₄)₂ (s) + H₂SO₄ (aq) → H₃PO₄ (aq) + CaSO₄ (aq)

Respuesta :

Answer:

The % yield is 64.38 %

Explanation:

Step 1: Data given

Mass of Ca3(PO4)2 = 26.8 grams

Mass of H2SO4 = 54.3 grams

Mass of H3PO4 formed = 10.9 grams

Molar mass of Ca3(PO4)2 =310.18 g/mol

Molar mass of H2SO4 = 98.08 g/mol

Molar mass of H3PO4 = 97.99 g/mol

Step 2: The balanced equation

Ca3(PO4)2 + 3 H2SO4 → 2 H3PO4 + 3 CaSO4

Step 3: Calculate moles Ca3(PO4)2

Moles Ca3(PO4)2 = mass Ca3(PO4)2 / molar mass Ca3(PO4)2

Moles Ca3(PO4)2 = 26.8 grams / 310.18 g/mol

Moles Ca3(PO4)2 = 0.0864 moles

Step 4: Calculate moles H2SO4

Moles H2SO4 = 54.3 grams / 98.08 g/mol

Moles H2SO4 = 0.554 moles

Step 5: Calculate limiting reactant

For 1 mol Ca3(PO4)2 we need 3 moles H2SO4 to produce 2 moles H3PO4 and 3 CaSO4

Ca3(PO4)2 is the limiting reactant. It will completely be consumed. (0.0864 moles).

H2SO4 is in excess. There will react 3*0.0864 = 0.2592 moles

There will remain: 0.554 - 0.2592 = 0.2948 moles

Step 6: Calculate moles H3PO4

For 1 mol Ca3(PO4)2 we need 3 moles H2SO4 to produce 2 moles H3PO4 and 3 CaSO4

For 0.0864 moles we'll have 2*0.0864 = 0.1728 moles H3PO4

Step 7: Calculate mass H3PO4

Mass H3PO4 = moles H3PO4 * molar mass H3PO4

Mass H3PO4 = 0.1728 moles * 97.99 g/mol

Mass H3PO4 = 16.93 grams

Step 8: Calculate percent yield

% yield = (actual yield / theoretical yield)*100%

% yield = (10.9/16.93)*100 %

% yield = 64.38 %

The % yield is 64.38 %

Oseni

The percentage yield if 10.9 g of H₃PO₄ is formed from the reaction would be 64.69%

From the balanced equation of the reaction:

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)

Mole ratio of Ca₃(PO₄)₂ to H₂SO₄ = 1:3

Mole of H₂SO₄ = mass/molar mass

                           = 54.3/98.08

                             = 0.55 mole

Mole of Ca₃(PO₄)₂ = 26.8/310.174

                               = 0.086 mole

Thus, Ca₃(PO₄)₂ is the limiting reactant.

Mole ratio of Ca₃(PO₄)₂ to H₃PO₄ = 1:2

Mole of H₃PO₄ = 0.086 x 2

                          = 0.172 mole

Theoretical yield (in g) of H₃PO₄ = mole x molar mass

                                    = 0.172 x 97.994

                                    = 16.85 g

Percentage yield of H₃PO₄ = actual yield/theoretical yield x 100

                                          = 10.9/16.85 x 100

                                          = 64.69%

More on percentage yield of reactions can be found here: https://brainly.com/question/20758645

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