The question is ...
One of the reactions that occur during baking is:
2NaHCO3 -> Na2CO3 + H2O + CO2
Suppose your recipe calls for 5 mL (approximately 3.0 g) of baking soda. What volume of carbon dioxide will be generated at 195 degrees celsius and 100 kPa to make the dough rise?
The answer is a volume of 0.69 L.
I actually have no idea how they got the answer. Can I use the combined gas law?
Can someone solve this problem step by step? Thanks so much in advance!
One of the reactions that occur during baking is:
2NaHCO3 -> Na2CO3 + H2O + CO2
Suppose your recipe calls for 5 mL (approximately 3.0 g) of baking soda. What volume of carbon dioxide will be generated at 195 degrees celsius and 100 kPa to make the dough rise?
The answer is a volume of 0.69 L.
I actually have no idea how they got the answer. Can I use the combined gas law?
Can someone solve this problem step by step? Thanks so much in advance!
-
I will try to suggest how to do it without giving the actual answer.
(Note that the calculated figures are deliberately a bit wrong.)
2NaHCO3 -> Na2CO3 + H2O + CO2
The equation tells you that 2 moles of sodium bicarbonate make
one mole of carbon dioxide. You also know that 1 mole of an
ideal gas, (carbon dioxide is near enough), occupy 22.4 litres
at standard pressure and temperature (s.t.p.)
1) Work out the weight in grams of 2 moles of sodium bicarbonate .
If you do not know how to do that by adding up atomic weights,
try this web page.
http://www.convertunits.com/molarmass/Na…
Lets say that was 400 g for those two moles. (It is not of course).
2) You only have 3 g. So now you do a calculation similar to
3/400 * 22.4 litres of carbon dioxide if it was released at s.t.p.
Suppose that was something like 0.15 litres. (It is not of course).
3) Do you know what standard pressure and temperature (s.t.p.) is ?
One definition is that it is at 273 absolute (or 0 °C) and 760 mm of Mercury.
But in case of unusual units:-
STP is 0 °C (32 °F or 273 Kelvin) and 1 atm (101.325 kPa, 14.7 PSI, 760 mmHg, or 760 Torr).
4) Let the volume that you have just calculated in 2) be v, and
let the volume of carbon dioxide generated at 195 degrees celsius and 100 kPa. be V
A key idea about the ideal gas laws is that :-
(Pressure * Volume)/Absolute Temperature is constant. So we may write
(101.325 kPa * v)/273 = (100 kPa. * V)/(273 +195)
Since you already have v it is just a calculator job to get V
Now you do have a good idea how they got the answer.
Go ahead and do it for yourself, and remember the method for exams.
Regards - Ian
(Note that the calculated figures are deliberately a bit wrong.)
2NaHCO3 -> Na2CO3 + H2O + CO2
The equation tells you that 2 moles of sodium bicarbonate make
one mole of carbon dioxide. You also know that 1 mole of an
ideal gas, (carbon dioxide is near enough), occupy 22.4 litres
at standard pressure and temperature (s.t.p.)
1) Work out the weight in grams of 2 moles of sodium bicarbonate .
If you do not know how to do that by adding up atomic weights,
try this web page.
http://www.convertunits.com/molarmass/Na…
Lets say that was 400 g for those two moles. (It is not of course).
2) You only have 3 g. So now you do a calculation similar to
3/400 * 22.4 litres of carbon dioxide if it was released at s.t.p.
Suppose that was something like 0.15 litres. (It is not of course).
3) Do you know what standard pressure and temperature (s.t.p.) is ?
One definition is that it is at 273 absolute (or 0 °C) and 760 mm of Mercury.
But in case of unusual units:-
STP is 0 °C (32 °F or 273 Kelvin) and 1 atm (101.325 kPa, 14.7 PSI, 760 mmHg, or 760 Torr).
4) Let the volume that you have just calculated in 2) be v, and
let the volume of carbon dioxide generated at 195 degrees celsius and 100 kPa. be V
A key idea about the ideal gas laws is that :-
(Pressure * Volume)/Absolute Temperature is constant. So we may write
(101.325 kPa * v)/273 = (100 kPa. * V)/(273 +195)
Since you already have v it is just a calculator job to get V
Now you do have a good idea how they got the answer.
Go ahead and do it for yourself, and remember the method for exams.
Regards - Ian