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qwelly [4]
2 years ago
8

A baker uses sodium hydrogen carbonate (baking soda) as the leavening agent in a banana-nut quick bread. the baking soda decompo

ses 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.
Chemistry
1 answer:
Kamila [148]2 years ago
6 0

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)

  • <u>Calculation for Reaction 1 :</u>

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

  • <u>Calculation for Reaction 2 :</u>

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


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