Theoretical Yield as a VOLUME
On this screen you will see how to calculate the theoretical yield of a product when the product is measured as a VOLUME.
What is the theoretical volume of Ethyne (measured at 0°C and 760 mm Hg) that can be produced from 5 grams of Calcium Carbide in the following reaction:
CaC2 + 2H2O ---------» C2H2 + Ca(OH)2
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Find the number of moles of Calcium Carbide used.
Moles = Mass/RMM
Moles of Calcium Carbide = 5/64.0 = 0.078125 moles RMM = 64.0
Find how many moles of Ethyne are produced.
From the balanced equation you can see that:
1 mole of Calcium Carbide would produce 1 mole of Ethyne. Thus:
0.078125 moles of Calcium Carbide would produce 0.078125 moles of Ethyne.
1 mole of Calcium Carbide would produce 1 mole of Ethyne. Thus:
0.078125 moles of Calcium Carbide would produce 0.078125 moles of Ethyne.
Finally, find the volume of Ethyne produced - as measured at 0°C and
760 mm Hg.
760 mm Hg.
At 0°C and 760 mm Hg, one mole of gas has a volume of 22.4 dm3
1 mole of Ethyne has a volume of 22.4 dm3
Thus 0.078125 moles of Ethyne have a volume of 0.078125/1 x 22.4 dm3
Thus 0.078125 moles of Ethyne have a volume of 1.75 dm3.
The theoretical yield of Ethyne is 1.75 dm3.
1 mole of Ethyne has a volume of 22.4 dm3
Thus 0.078125 moles of Ethyne have a volume of 0.078125/1 x 22.4 dm3
Thus 0.078125 moles of Ethyne have a volume of 1.75 dm3.
The theoretical yield of Ethyne is 1.75 dm3.