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Thu 9 Apr, 2020 10:27 am
1(a)If 0.1 mole of benzoic acid (C6H5COOH) was reacted with oxygen gas,calculate the volume of carbon dioxide to be produced at room temperature and pressure (r.t.p).
(b)If the volume of carbon dioxide obtained was 6.5dm^3,what was the percentage yield of the carbon dioxide?
@Samuel Kashina,
Part a: write down the expected reaction. How many moles of CO2 would you expect from 0.1 mole of benzoic acid? At STP, one mole of gas is approximately 22.4 liters in volume, so you can get the expected volume.
Part b: What percentage is the given volume compared to the expected volume?
It's RTP not s.t.p
V=24dm³
The equation is like this
C6H5COOH+O2--------------->7CO2+H2O
Help me please
@Samuel Kashina,
Yes kind of. You could balance the amount of Oxygen and water, but you got the important part, 1 mole of benzoic acid produces seven moles of CO2.
So how do we calculate the volume,since
Concentration=Mass/volume or concentration=no.of mols×volume
Which formula do we use and what stages do we take to arrive at the answer.
@engineer,
engineer wrote:
Part a: write down the expected reaction. How many moles of CO2 would you expect from 0.1 mole of benzoic acid? At STP, one mole of gas is approximately 22.4 liters in volume, so you can get the expected volume.
Part b: What percentage is the given volume compared to the expected volume?
a) So you get 7 moles per 1 that you start with and you started with 0.1 moles so post reaction you will have ... 0.7 moles. Each mole is 22dm3 at RTP so the total volume is 0.7 x 22 = 15.4 dm3. Do you see where that comes from?
b) You expected 15.4 but only got 6.5. What percentage is that?
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