Question

Type the correct answer in the box. Express your answer to three significant figures.

Answer 1

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 2

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

Explanation:

$3Ca(NO_3)_2+2Na_3PO_4\rightarrow 6NaNO_3+Ca_3(PO_4)_2$

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

According to reaction:

3 mole of $Ca(NO_3)_2$ gives 1 mole of $Ca_3(PO_4)_2$

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

Mass of $Ca_3(PO_4)_2$: number of Moles × Molar mass

$=0.1933\times 310.18 g/mol=59.95 g$

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