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` =
`(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
`(1mole)/(2moles)* 0.0119moles=0.00595moles` of
`CO_2`

Using ideal gas equation :

`PV=nRT\\V=(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=((0.00595mole)* (0.0821Latm/moleK)* 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=(nRT)/(P)`

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

`V=((0.0119mole)* (0.0821Latm/moleK)* 443K)/(0.900atm)=0.48089L=480.89ml`

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