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Calcium nitrate reacts with sodium phosphate to form sodium nitrate and calcium phosphate. Given 96.1 grams of calcium nitrate, what’s the theoretical yield of calcium phosphate? Use the periodic table and the polyatomic ion resource.

The mass of calcium phosphate the reaction can produce is

grams.

Respuesta :

Given:

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

To determine:

Theoretical yield of calcium phosphate, Ca3(PO4)2

Explanation:

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 =  96.1 g

Molar mass of Ca(NO3)2 =  164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles

Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g

Answer: The mass of calcium phosphate the reaction can produce is  59.95 g.

Explanation:

[tex]3Ca(NO_3)_2+2Na_3PO_4\rightarrow 6NaNO_3+Ca_3(PO_4)_2[/tex]

Moles of [tex]Ca(NO_3)_2=\frac{\text{given Mass of }Ca(NO_3)_2}{\text{Molar Mass of }Ca(NO_3)_2}=\frac{96.1 g}{164.08 g/mol}=0.58 mol[/tex]

According to reaction:

3 mole of [tex]Ca(NO_3)_2[/tex] gives 1 mole of [tex]Ca_3(PO_4)_2[/tex]

Then 0.58 moles of [tex]Ca(NO_3)_2[/tex] will give = [tex]\frac{1}{3}\times 0.58[/tex] moles of [tex]Ca_3(PO_4)_2[/tex] that is 0.1933 moles

Mass of [tex]Ca_3(PO_4)_2[/tex]: number of Moles × Molar mass

[tex]=0.1933\times 310.18 g/mol=59.95 g[/tex]

The mass of calcium phosphate the reaction can produce is  59.95 g.

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