Question

A baker uses sodium hydrogen carbonate (baking soda) as the leavening agent in a banana-nut quickbread. The baking soda decomposes according to two possible reactions. Reaction 1: 2 NaHCO3(s)-Na2CO3(s) + Њ0() + CO2(g) Reaction 2: NaHCO3(s) H(a) H00) CO2(g) +Na (aq) Calculate the volume (in mL) of Co2 that forms at 220.°C and 0.995 atm per gram of NaHCO3 by each of the reaction processes mL (Reaction 1) mL (Reaction 2)

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Answer 1

Reaction 1: From the balanced chemical equation, 2 mol NaHCO₃ give 1 mol CO₂.

We know that at STP (0°C and 1 atm) 1 mol of a gas has volume V₁=22.4 L

Temperature T_{​​​​​​1}=0°C=0+273 K=273 K

Pressure P_{​​​​​1}=1 atm

Number of moles n_{​​​​​​1}​​​​​=1 mol

Molar mass of NaHCO₃=Molar mass of Na+Molar mass of H+Molar mass of C+3xmolar mass of O=23 g/mol+12 g/mol+1 g/mol+3x16 g/mol=36 g/mol+48 g/mol=84 g/mol

Number of moles in 1 g NaHCO₃=mass/molar mass=1 g/84 g/mol=0.012 mol

As 2 mol NaHCO₃ gives 1 mol CO₂

So 1 mol NaHCO₃ gives 1/2 mol CO₂

0.012 mol NaHCO₃ gives (1/2)x0.012 mol=0.006 mol CO₂

So n₂=0.006 mol

At temperature T₂=220°C+273K=493 K

Pressure P₂=0.995 atm

Also we know that PV=nRT

So R=PV/nT

(Where R=Gas constant, P=Pressure, V=Volume, n=Number of moles, T=Temperature in K)

So applying $\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}$ (=R)....(1) and substituting values in the equation

1atm × 22.4L 0.99-atm × V2 1mol × 273K 0.0067701 × 493K

So V₂=Volume==0.24393 L x 1000 mL/L

1atm × 22.4L × 0.00677101 × 493K 1mol × 273K × 0.995atm

=243.93 mL (1 L=1000 mL)

So volume of CO₂ from reaction 1=243.93 mL

Reaction 2: 1 mol NaHCO₃ gives 1 mol CO₂

0.012 mol NaHCO₃ gives 0.012 mol NaHCO₃

So for reaction 2, n₂=0.012 mol

So using the same remaining values of P₁, P₂, n₁, V₁, T₁ and T₂ as from reaction 1 in equation (1)

1atm × 22.4L 0.99-atm × V2 1mol × 273K 0.01277201 × 493K

V₂==0.48785 L x 1000 mL/L

latm × 22.4L × 0.0127701 × 493K 1mol × 273K × 0.995atm

=487.85 mL

So volume of CO₂ from reaction 2=487.85 mL