5 – 7. Answer each of the questions in order to answer the final question. Given

Question

5 – 7. Answer each of the questions in order to answer the final question.

Given the following balanced chemical equation and the information about each of the reactants, calculate the grams of Ca3(PO4)2 that will be formed, assuming 100% reaction.

2 H3PO4 + 3 Ca(NO3)2 à Ca3(PO4)2 + 6 HNO3

Molar masses =          98.0 g 164.0g             310.0 g            63.0 g

Grams/mL =               15.0 mL           22.0 g

Density =                     1.88 g/mL        solid                 solid

5. Moles of Ca3(PO4)2 formed from 15.0 mL of phosphoric acid?

a. 0.144        b. 782    c. 0.0814     d. 0.288

6. Moles of calcium phosphate made from 22.0 grams of calcium nitrate?

a. 0.0447         b. 0.134           c. 0.402           d. 13.9

7. Grams of calcium phosphate formed?

a. 13.9             b. 41.5            c. 124.6           d. 44.64

Explanation / Answer

H3PO4 mass = density x volume = 1.88 x 15 = 28.2 g

H3PO4 moles = mass of H3PO4 / molar mass of H3PO4 = 28.2 /98 = 0.287755

Ca3(PO4)2 moles = 1/2 H3PO4 moles = 0.287755/2 = 0.143877 = 0.144 ,

6) Ca(NO3)2 moles = 22 /164 = 0.134146

as per reaction Ca3(PO4)2 moles =   1/3 ( Ca(NO3)2 = 1/3( 0.134146) = 0.0447

7) 2H3Po4 reacts with 3 Ca3(PO4)2

hence for 0.287755 moles H3PO4 Ca3(PO4)2 moles expected = ( 3/2) 0.287755   = 0.43164

but we had 0.134146 moles of Ca(NO3)2 , hence ca(NO3)2 is limiting reagent , and ca3(PO4)2 moles depend on Ca(NO3)2   = 0.0447

Ca3(PO4)2 mass = moles of it x molar mass of Ca3(PO4)2 = 0.0447 x 310 = 13.9 g

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