contestada

Suppose the reaction Ca3(PO4)2 + 3H2SO4 → 3CaSO4 + 2H3PO4 is carried out starting with 149 g of Ca3(PO4)2 and 86.9 g of H2SO4. How much phosphoric acid will be produced?

Respuesta :

Answer:

57.9 grams .

Explanation:

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

1 mole         3 moles                              2 moles

149 g of Ca₃(PO₄)₂ = 149 / 310 = .48 moles

86.9 g of H₂SO₄ = 86.9 / 98 = .8867 moles

Here H₂SO₄  is the limiting reagent .

.8867 / 3 = .2955

.2955 moles of Ca₃(PO₄)₂ reacts with  .8867 mole of H₂SO₄  to give

2 x .2955 moles of H₃PO₄

H₃PO₄ produced = 2 x .2955 moles

= .591 moles

= .591 x 98 = 57.9 grams .

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