Question

1. How many moles of Ca(NO₃)₂*4H₂O are in 50.0 mL of 0.0906 M Ca(NO₃)₂*4H₂O? Report your answer with 3 significant figures. Do not include units.

2. How many moles of Na₃PO₄*12H₂O are in 50.0 mL of 0.0550 M Na₃PO₄*12H₂O? Report your answer with 3 significant figures. Do not include units.

3. If 50.0 mL of 0.0906 M Ca(NO₃)₂*4H₂O are reacted with 50.0 mL of 0.0550 M Na₃PO₄*12H₂O, using the balanced chemical equation from the video, determine which is the limiting reactant.

4. What is the theoretical yield of Ca₃(PO₄)₂ (in grams)? Report your answer with 3 significant figures. Do not include units.

5.  If the actual yield of Ca₃(PO₄)₂ were 0.40 g and the theoretical yield of Ca₃(PO₄)₂ were 0.41 g, calculate the percent yield of Ca₃(PO₄)₂. Report your answer with 2 significant figures. Do notinclude units.

Answer 1

1. Moles of Ca(NO3)2.4H2O in 50.0 mL of 0.0906 M Ca(NO3)2.4H2O = (volume of the solution in L)*[molarity of Ca(NO3)2.4H2O]

= (50.0 mL)*(0.0906 M)

= (50.0 mL)*(1 L)/(1000 mL)*(0.0906 M)

= 0.00453 (ans).

2. Moles of Na3PO4.12H2O in 50.0 mL of 0.0550 M Na3PO4.12H2O = (volume of the solution in L)*[molarity of Na3PO4.12H2O]

= (50.0 mL)*(0.0550 M)

= 0.00275 (ans).

3. The balanced chemical equation for the reaction between Ca(NO3)2.4H2O and Na3PO4.12H2O is given as

3 Ca(NO3)2.4H2O (aq) + 2 Na3PO4.12H2O (aq) ----------> Ca3(PO4)2 (g) + 6 NaNO3 (aq) + 36 H2O (l)

As per the stoichiometric equation,

3 moles Ca(NO3)2.4H2O = 2 moles Na3PO4.12H2O

Determine the limiting reactant.

0.00453 mols Ca(NO3)2.4H2O will react with

[0.00453 mol Ca(NO3)2.4H2O]*[2 moles Na3PO4.12H2O]/[3 moles Ca(NO3)2.4H2O]

= 0.00302 mols Na3PO4.12H2O

0.00275 mole Na3PO4.12H2O will react with

[0.00275 mol Na3PO4.12H2O]*[3 moles Ca(NO3)2.4H2O]/[2 moles Na3PO4.12H2O]

= 0.004125 mol Ca(NO3)2.4H2O.

The reaction system contains 0.00275 mol Na3PO4.12H2O which is less than what is required to react with 0.00453 mol Ca(NO3)2.4H2O. Hence, Na3PO4.12H2O is the limiting reactant.

4. As per the equation given above,

2 mols Na3PO4.12H2O = 1 mol Ca3(PO4)2.

Therefore,

0.00275 mol Na3PO4.12H2O = [0.00275 mol Na3PO4.12H2O]*[1 mol Ca3(PO4)2]/[2 moles Na3PO4.12H2O]

= 0.001375 mol Ca3(PO4)2.

The atomic masses are

Ca: 40 g/mol

P: 31 g/mol

O: 16 g/mol

Molar mass of Ca3(PO4)2 = (3*40 + 2*31 + 8*16) g/mol = 310 g/mol

Grams of Ca3(PO4)2 obtained = (0.001375 mol)*(310 g/mol)

= 0.42625 g

≈ 0.426 g

The grams of Ca3(PO4)2 obtained = 0.426 (ans).

5. The theoretical yield is assumed to be 0.41 g and the actual yield is 0.40 g.

Percent yield = (0.40 g)/(0.41 g)*100

= 97.56

≈ 98. (ans).