How many grams of the solid can be formed from 206 g of calcium nitrate and 148 g of phosphoric acid? Answer in units of g.

Answer 1

First, balance the reaction: 3 Ca (NO₃)₂ + 2 H₃PO₄ --> Ca₃ (PO₄)₂ + 6 HNO₃.

From 206 g calcium nitrate at 164.1 g/mol we have 1.255 mol calcium nitrate * (1 mol calcium phosphate/ 3 mol calcium nitrate) = 0.4184 mol calcium phosphate (the precipitate). Doing the same stoich with 148 g phosphoric acid produces more calcium phosphate than this, so the calcium nitrate is limiting.

Finally, (0.4184 mol calcium phosphate) * (310.3 g/mol) = 130 g calcium phosphate.

Answer 2

Balanced equation...

3Ca(NO₃)₂(aq) + 2H₃PO₄(aq) ==> Ca₃(PO₄)₂(s) + 6 HNO₃(aq)

206 g.....................148 g....................?

1.25 moles............1.51 moles............?

Since calcium nitrate is limiting, the moles of calcium phosphate that can form is...

1.25 moles calcium nitrate x 1 mol calcium phosphate/3 mol calcium nitrate = 0.417 moles solid product

mass of solid = 0.417 moles x 310 g/mol = 129 g