A mixture of CO and CO2 having a volume of 20 mL is mixed with x mL of oxygen and electrically sparked. The volume after explosion is (16+x) mL under the same conditions. What would be the residual volume if 30 mL of the original mixture is treated with aqueous NaOH
2 answers
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This is the craziest problem I've seen. My solution may be even crazier but here goes.
mL CO + mL CO2 = 20 mL
mL CO + mL CO2 + xmLO2 = 20 + x mL total
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RXNS: 2CO + O2 ==> 2CO2
Here is my reasoning: CO2 and O2 will not react; therefore, the original amount of CO2 will be unchanged between spark time and no spark time.
The reaction between CO and O2 is above.For every mole of CO initially that reacts with O2 we will get an equal amount of CO2; therefore, loss of 1 mL CO will produce 1 mL of CO2 so that volume does not change either.
Therefore, I think that the 4 mL lost [(20 mL + x) - (16 mL + x)] is due to the O2 being used. That came from 8 mL CO that was oxidized so the amount of CO2 in the original container is 20- 8 = 12 mL. In a 30 mL sample of the same gas that would be 12 x 30/20 = 18 mL CO2. If all of the CO2 reacts with NaOH [NaOH(s) + CO2(g) ==> Na2CO3(s) + H2O(l) that will leave CO gas only from the mixture. 30 mL gas total - 18 mL CO2 = 12 mL residual. The flaws I see in this is that no where in the problem does it say that the x mL O2 added is an excess and that not all of the CO is oxidized. Anyone, please let me know if you have different thoughts. If Anonymous gets the answers please let me know.
mL CO + mL CO2 = 20 mL
mL CO + mL CO2 + xmLO2 = 20 + x mL total
--------------------------------------------------------
RXNS: 2CO + O2 ==> 2CO2
Here is my reasoning: CO2 and O2 will not react; therefore, the original amount of CO2 will be unchanged between spark time and no spark time.
The reaction between CO and O2 is above.For every mole of CO initially that reacts with O2 we will get an equal amount of CO2; therefore, loss of 1 mL CO will produce 1 mL of CO2 so that volume does not change either.
Therefore, I think that the 4 mL lost [(20 mL + x) - (16 mL + x)] is due to the O2 being used. That came from 8 mL CO that was oxidized so the amount of CO2 in the original container is 20- 8 = 12 mL. In a 30 mL sample of the same gas that would be 12 x 30/20 = 18 mL CO2. If all of the CO2 reacts with NaOH [NaOH(s) + CO2(g) ==> Na2CO3(s) + H2O(l) that will leave CO gas only from the mixture. 30 mL gas total - 18 mL CO2 = 12 mL residual. The flaws I see in this is that no where in the problem does it say that the x mL O2 added is an excess and that not all of the CO is oxidized. Anyone, please let me know if you have different thoughts. If Anonymous gets the answers please let me know.
1 answer