Respuesta :
Answer:
51.1 g of Ca₃(PO₄)₂(s) can be made in this reaction
Explanation:
The reactans are CaCl₂ and Na₃PO₄. Let's determine the reaction:
3CaCl₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) ↓ + 6NaCl(aq)
We convert the mass of chloride to moles:
55 g . 1 mol/ 110.98 g = 0.495 moles
Ratio is 3:1. Let's make a rule of three to find the answer in moles:
3 moles of chloride can produce 1 mol of phosphate
Therefore 0.495 moles will produce (0.495 . 1) / 3 = 0.165 moles
We convert the moles to mass:
0.165 mol . 310.18 g /1 mol = 51.1 g
The mass of calcium phosphate produced has been 51.1 g.
The balanced chemical equation of for the reaction has been:
[tex]\rm 3\;CaCl_2\;+\;2\;Na_3PO_4\;\rightarrow\;Ca_3(PO_4)_2\;+\;6\;NaCl[/tex]
The balanced chemical equation has been given that 3 moles of calcium chloride produces 1 moles of calcium phosphate.
The moles of calcium chloride in 55 g sample has been given as:
[tex]\rm Moles=\dfrac{Mass}{Molar\;mass}[/tex]
Substituting the values:
[tex]\rm Moles=\dfrac{55}{110.98}\\Moles\;CaCl_2=0.495\;mol[/tex]
The moles of calcium chloride has been 0.495 mol.
The moles of calcium formed has been given by:
[tex]\rm 3\;mol\;CaCl_2=1\;mol\;Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=\dfrac{1}{3}\;\times\;0.495\;mol\; Ca_3(PO_4)_2\\0.495\;mol\;CaCl_2=0.165\;mol\;Ca_3(PO_4)_2\\[/tex]
The moles of calcium phosphate formed has been 0.165 mol.
The mass of calcium phosphate has been:
[tex]\rm Mass=Moles\;\times\;molar\;mass\\Mass=0.165\;\times\;310.18\;g\\Mass=51.1\;g[/tex]
The mass of calcium phosphate produced has been 51.1 g.
For more information about moles formed, refer to the link:
https://brainly.com/question/4066080
