Question

Calcium dihydrogen phosphate, Ca(H2PO4)2, and sodium hydrogen carbonate, NaHCO3, are ingredients of baking powder that react with each other to produce CO2, which causes dough or batter to rise:

Ca(H₂PO₄)₂(s) + NaHCO₃(s) → CO₂(g) + H₂O(g) + CaHPO₄(s) + Na₂HPO₄(s)[unbalanced]

If the baking powder contains 31.0% NaHCO₃ and 35.0% Ca(H₂PO₄)₂ by mass:

(a) How many moles of CO₂ are produced from 1.59 g of baking powder?

mol CO₂ =

(b) If 1 mol of CO₂ occupies 37.0 L at 350°F (a typical baking temperature), what volume of CO₂ is produced from 1.59 g of baking powder?

L=

Answer 1

balanced reaction is

Ca(H2PO4)2 + 2NaHCO3 --> 2CO2 + H2O + CaHPO4 + Na2HPO4

Mass of Ca(H2PO4)2 = ( 35/100) x 1.59 = 0.5565 g
Moles of Ca(H2PO4)2 = mass/ molar mass = 0.5565 /234.05 = 0.002378

Mass of NaHCO3 = ( 31/100) x 1.59 = 0.4929 g

Moles of NaHCO3 = ( 0.4929/84) = 0.00587

as per reaction NaHCO3 and Ca(H2PO4)2 react in 2:1 , hence for 0.00587 moles NaHCo3 we need ( 1/2) 0.00587 = 0.00294 moles Ca(H2PO4)2 moles but we had only 0.002378 moles Ca(H2PO4)2

hence Ca(H2PO4)2 is limiting reagent ,

CO2 moles = 2 x Ca(H2PO4)2 moles = 2 x 0.002378 = 0.004756

b) 1 mol occupies 37 L

0.004756 moles CO2 occupies = 37 x 0.004756 = 0.176 L