stoiciometry and moles

Question

Starting with the following equation, Ca₃(PO₄)₂(s) + H₂SO₄(aq) → CaSO₄(s) + H₃PO₄(aq) calculate the mass in grams of Ca₃(PO₄)₂ that will be required to produce 475 grams of CaSO₄.

Answer 1

Ca₃(PO₄)₂ (s) + H₂SO₄ (aq) → CaSO₄ (s) + H₃PO₄ (aq)

Balance the equation:

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 3CaSO₄ (s) + 2H₃PO₄ (aq)

Starting with 475 grams of CaSO₄, calculate the number of moles using the molar mass:

Molar mass is:

Ca: 40.078(1) = 40.078 g/mol

S: 32.066(1) = 32.066 g/mol

O: 15.999(4) = 63.996 g/mol

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CaSO₄ = 136.14 g/mol

Calculate the number of moles by dividing the mass by the molar mass: (475 g)/(136.14 g/mol) = 3.489 moles

Looking at the balanced chemical reaction equation, it takes 1 mole of Ca₃(PO₄)₂ to make 3 moles of CaSO₄ therefore, if 3.489 moles of CaSO₄ was created, we must have started with 3.489/3 or 1.163 moles of Ca₃(PO₄)₂.

Now, to find the mass of Ca₃(PO₄)₂ we must again use the molar mass.

Calculate the molar mass of Ca₃(PO₄)₂:

Ca: 40.078(3) = 120.234 g/mol

P: 30.974(2) = 61.948 g/mol

O: 15.999(8) = 127.992 g/mol

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Ca₃(PO₄)₂ = 310.174 g/mol

This time multiply the molar mass by the number of moles to get the mass:

(310.174 g/mol)(1.163 mol) = 360.738 g

Round the appropriate number of sig figs: 361 g of Ca₃(PO₄)₂ was required