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Question

How many grams of calcium phosphate (precipitate) could be produced by combining 12.354g of calcium nitrate dissolved in water with 16.076g of rubidium phosphate dissolved in water? Report your answer to 3 decimal place

Answer 1

First, write the correctly balanced equation for the reaction taking place.

3Ca(NO₃)₂(aq) + 2Rb₃PO₄(aq) ==> Ca₃(PO₄)₂(s) + 6RbNO₃(aq) ... balanced equation

Next, we'll find which reactant is in limiting supply. An easy way to do this is to simply divide the mols of each reactant by its coefficient in the balanced equation.

For Ca(NO₃)₂: 12.354 g Ca(NO₃)₂ x 1 mol / 164.09 g = 0.07529 mols (÷3->0.0251)

For Rb₃PO₄: 16.076 g Rb₃PO₄ x 1 mol / 351.38 g = 0.04575 mols (÷2->0.0229)

Therefore, since 0.045 is less than 0.075, Rb₃PO₄ is the limiting reactant and will dictate how much product is formed. We will use the moles of Rb₃PO₄ to find the moles of Ca3(PO4)2 and then convert that to grams.

0.04575 mols Rb₃PO₄ x 1 mol Ca₃(PO₄)₂ / 3 mols Rb₃PO₄ = 0.01525 mols Ca₃(PO₄)₂ formed

0.01525 mols Ca₃(PO₄)₂ x 310.18 g / mol = 4.370 g Ca₃(PO₄)₂ formed