Apiece of sodium metal reacts completely with water as follows: 2na(s) + 2h₂o(l) 2naoh(aq) + h₂(g) the hydrogen gas generated is collected over water at 23.0°c. the volume of the gas is 397 ml measured at 1.013 atm. calculate the number of grams of sodium used in the reaction. (the vapor pressure of water at 23.0°c = 0.0277 atm.)
Firstly, we can use the general gas equation to get the volume produced at s.t.p.
The form of the general gas equation used is :
P1V1/T1 = P2V2/T2
The values of each are given below :
P1 = (1.013 - 0.0277) = 0.9853atm.
This subtraction is because the gas was collected over water and hence there is bound to be contribution of water to the total pressure.
T1 = 23 + 273 = 296K
This is the conversion from celcius temperature scale to absolute temperature scale. This is the scale used in this kind of calculations.
T2 = standard temperature = 273k
P2 = standard pressure = 1atm.
V2 = ?
Substituting the values will yield:
(0.9853 × 397)/296 = (1 × V2)/273
Via calculations, V2 = 360.77ml
Now we know the volume produced at S.t.p
Secondly, we use a relation to get the number of moles yielded.
At s.t.p, 1 mole of a gas occupies a volume of 22,400ml.
If 1 mole occupies 22,400ml volume, x moles will occupy a volume of 360.77ml
Hence, the number of moles is obtained by dividing 360.77ml by 22,400ml = 0.016mole
Now, we go back to the reaction equation.
We can see that 2 moles of sodium metal yielded 1 mole of hydrogen gas.
Hence, y moles of sodium metal would have yielded 0.016mole of hydrogen gas.
Thus,y equals 2 × 0.016 = 0.032moles
Or we just simply say they have a mole ratio of 2 to 1.
Lastly, we can now get the mass of the sodium metal used in the reaction.
We use the relation as follows,
Mass of sodium metal used = number of moles of sodium metal × atomic mass of sodium metal
The atomic mass of sodium metal = 23
Hence, the mass used = 0.032 × 23 = 0.736g
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