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How many grams of solid Ca3(PO4)2 can be formed from a reaction of 51.4 mL of 0.758 M CaCl2 and 21.5 mL of 1.04 M Na3PO4?

Respuesta :

Dejanras
Answer is: 3,41 g of Ca₃(PO₄)₂.
Chemical reaction: 3CaCl₂ + 2Na₃PO₄ → Ca₃(PO₄)₂ + 6NaCl.
V(CaCl₂) = 51,4 mL = 0,0514 L.
c(CaCl₂) = 0,758 mol/L.
n(CaCl₂) = V(CaCl₂) · c(CaCl₂).
n(CaCl₂) = 0,0514 L · 0,758 mol/L.
n(CaCl₂) = 0,039 mol.
V(Na₃PO₄) = 21,5 mL = 0,0215 L.
c(Na₃PO₄) = 1,04 mol/L.
n(Na₃PO₄) = 0,0215 L · 1,04 mol/L.
n(Na₃PO₄) = 0,022 mol, limiting reagens.
From chemical reaction: n(Na₃PO₄) : n(Ca₃(PO₄)₂) = 2 : 1.
n(Ca₃(PO₄)₂) = 0,011 mol.
m(Ca₃(PO₄)₂) = 0,011 mol · 310,17 g/mol.
m(Ca₃(PO₄)₂) = 3,41 g.
Oseni

The amount of Ca3(PO4)2 that would be formed from the reaction will be 10.40 g

Stoichiometric reaction

From the equation of the reaction:

3CaCl2 + 2Na3PO4 = Ca3(PO4)2 + 6NaCl

Mole of 51.4 mL, 0.758 M CaCl2 = 51.4/1000 x 0.758

                                                        = 0.039 moles

Mole of 21.5 mL, 1.04 M Na3PO4 = 21.5/1000 x 1.04

                                                          = 0.022 moles

Mole ratio of CaCl2 and Na3PO4 = 3:2

Thus, Na3PO4 appears to be limiting.

Mole ratio of Na3PO4 and Ca3(PO4)2 = 2:1

Equivalent mole of Ca3(PO4)2 = 0.033 moles

Mass of 0.033 moles  Ca3(PO4)2 = 0.033 x 310.18

                                                     = 10.40 g

More on stoichiometric calculations can be found here: https://brainly.com/question/8062886

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