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For the equation 2H₃PO₄ + 3Ca ⇒ 3H₂ + Ca₃(PO₄)₂ suppose you had 27.4 grams of H₃PO₄?
a. How many grams of Ca would you need to react with this number of moles of H₃PO₄?

b. How many grams of hydrogen gas will be produced?

c. How many moles of calcium phosphate will be produced?

Please answer all questions clearly, with work shown! Thank you!!

Respuesta :

jpelezo1

Answer:

Explanation:

Given 2H₃PO₄ + 3Ca ⇒ 3H₂ + Ca₃(PO₄)₂

          27.4g

       =(27.4g/98g/mol)

       = 0.280 mole H₃PO₄ used

Ca Used:

Rxn Ratio H₃PO₄ : Ca => 2:3

∴ moles Ca used = 3/2(0.280) mole Ca =0.420 mole Ca

grams Ca used = 0.420 mole Ca x 40 g/mole = 16.8 grams Ca used

Grams H₂ Produced:

Rxn Ratio H₃PO₄ : H₂ => 2:3 => moles H₂ produced = 3/2(0.28 mole H₂) = 0.420 mole H₂ = (0.420 mole H₂ x 2.02 g/mole) = 0.8484 grams H₂(g) Produced.

Moles Ca₃(PO₄)₂ Produced:

FWt Ca₃(PO₄)₂ = 310 g/mole

Rxn Ratio H₃PO₄ : Ca₃(PO₄)₂  => 2:1

∴moles Ca₃(PO₄)₂ Produced = 1/2(moles H₃PO₄ used)

= 1/2(0.28 mole) = 0.14 mole Ca₃(PO₄)₂  

If you want grams Ca₃(PO₄)₂, multiply moles Ca₃(PO₄)₂ by formula weight.

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