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 3.50 g of baking powder?

Answer 1

0.012 mol of CO₂

Step-by-step explanation:

The balanced equation is

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

On 3.50 g of baking powerd:

mCa(H₂PO₄)₂ = 0.35*3.50 = 1.225 g

mNaHCO₃ = 0.31*3.50 = 1.085 g

The molar masses are: Ca = 40 g/mol; H = 1 g/mol; P = 31 g/mol; O = 16 g/mol; Na = 23 g/mol; C = 12 g/mol. So:

Ca(H₂PO₄)₂: 40 + 4x1 + 31 + 8x16 = 203 g/mol

NaHCO₃: 23 + 1 + 12 + 3x16 = 84 g/mol

The number of moles is the mass divided by molar mass, so:

nCa(H₂PO₄)₂ = 1.225/203 = 0.006 mol

nNaHCO₃ = 1.085/84 = 0.0129 mol

First, let's find which reactant is limiting. Testing for Ca(H₂PO₄)₂, the stoichiometry is:

1 mol of Ca(H₂PO₄)₂ ---------- 2 mol of NaHCO₃

0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol

So, NaHCO₃ is in excess. The stoichiometry calculus must be done with the limiting reactant, then:

1 mol of Ca(H₂PO₄)₂ ------------- 2 mol of CO₂

0.006 of Ca(H₂PO₄)₂ -------- x

By a simple direct three rule:

x = 0.012 mol of CO₂