Question

A baker uses sodium hydrogen carbonate (baking soda) as the leavening agent in a banana-nut quick bread. the baking soda decomposes according to two possible reactions. reaction 1: 2 nahco3(s) → na2co3(s) + h2o(l) + co2(g) reaction 2: nahco3(s) + h+(aq) → h2o(l) + co2(g) + na+(aq) calculate the volume (in ml) of co2 that forms at 170.°c and 0.900 atm per gram of nahco3 by each of the reaction processes.

Answer 1

Answer : The volume of $CO_2$ in Reaction 1 = 240.45 ml

The volume of $CO_2$ in Reaction 2 = 480.89 ml

Solution : Given,

Temperature = $170^oC=170+273=443K$         ($1^oC=273K$)

Pressure = 0.900 atm

The mass of $NaHCO_3$ = 1 gram

Molar mass of $NaHCO_3$ = 84.007 g/mole

The given reactions are,

Reaction 1 : $2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(l)+CO_2(g)$

Reaction 2 : $NaHCO_3(s)+H^+(aq)\rightarrow H_2O(l)+CO_2(g)+Na^+(aq)$

• Calculation for Reaction 1 :

First we have to calculate the moles of $NaHCO_3$.

Moles of $NaHCO_3$ = $\frac{1g}{84.007g/mole}=0.0119moles$

From the Reaction 1, we conclude that

2 moles of $NaHCO_3$ gives 1 mole of $CO_2$

0.0119 moles of $NaHCO_3$ gives $\frac{1mole}{2moles}\times 0.0119moles=0.00595moles$ of $CO_2$

Using ideal gas equation :  

$PV=nRT\\V=\frac{nRT}{P}$

where,

P = pressure of gas

V = volume of gas

n = Number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K

Now put all the given values in ideal gas law, we get the volume of $CO_2$

$V=\frac{(0.00595mole)\times (0.0821Latm/moleK)\times 443K}{0.900atm}=0.24045L=240.45ml$

The volume of $CO_2$ in Reaction 1 is 240.45 ml

• Calculation for Reaction 2 :

The moles of $NaHCO_3$ = 0.0119 moles

From the Reaction 2, we conclude that

1 moles of $NaHCO_3$ gives 1 mole of $CO_2$

So, the moles of $CO_2$ = the moles of $NaHCO_3$ = 0.0119 moles

Using ideal gas equation,

$PV=nRT\\V=\frac{nRT}{P}$

Now put all the given values in ideal gas law, we get the volume of $CO_2$

$V=\frac{(0.0119mole)\times (0.0821Latm/moleK)\times 443K}{0.900atm}=0.48089L=480.89ml$

The volume of $CO_2$ in Reaction 2 is 480.89 ml