Question

Be sure to answer all parts.

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 5.98 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 5.98 g of baking powder?

__L

Answer 1

Answer 2

Answer:

(a) Moles of CO₂ Produced:

1. Calculate masses of reactants in 5.98 g baking powder:

• NaHCO₃: $5.98\text{g}\times 0.310=1.8538\text{g}$

• Ca(H₂PO₄)₂: $5.98\text{g}\times 0.350=2.093\text{g}$

2. Convert masses to moles:

• Moles of NaHCO₃: $\frac{1.8538}{84.01}\approx 0.0221\text{mol}$

• Moles of Ca(H₂PO₄)₂: $\frac{2.093}{234.05}\approx 0.00894\text{mol}$

3. Determine limiting reactant (balanced reaction: 1:2 ratio):

• NaHCO₃ is limiting (0.0221 mol vs. required 0.0179 mol for Ca(H₂PO₄)₂).

4. Moles of CO₂ produced (1:1 ratio with NaHCO₃):

   $$\boxed{{\displaystyle 0.0221{\text{mol CO}}_2}}$$

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(b) Volume of CO₂ at 350°F:

1. Use given volume per mole (37.0 L/mol):

   $$0.0221\text{mol}\times 37.0\text{L/mol}=\boxed{{\displaystyle 0.818\text{L}}}$$

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Key Notes:

• Balanced reaction: Assumed ${\text{Ca(H}}_2{\text{PO}}_4{\text{)}}_2+2{\text{NaHCO}}_3\to 2{\text{CO}}_2+{\text{H}}_2\text{O}+{\text{CaHPO}}_4+{\text{Na}}_2{\text{HPO}}_4$.

• Limiting reactant: NaHCO₃ dictates CO₂ yield.

• Volume calculation: Direct scaling from moles to given conditions