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
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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.
To find the residual volume, we can set up an equation using the concept of stoichiometry.
Let's consider the reaction of CO and CO2 with oxygen:
2CO + O2 -> 2CO2 (1)
Given that the initial volume of the mixture containing CO and CO2 is 20 mL, and it becomes (16+x) mL after the explosion. This means that 4 mL of gas was consumed during the reaction.
Now, let's find the volume of oxygen required to react with the CO and CO2 mixture:
According to equation (1), 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.
Since the volume is directly proportional to the number of moles of gas, we can represent this as:
2CO + O2 -> 2CO2
2 volumes + x volumes -> 2 volumes
Therefore, the 20 mL mixture reacts with half of the required oxygen volume, which is 2/3 * 4 mL = (8/3) mL.
Let's now determine the residual volume when 30 mL of the original mixture is treated with aqueous NaOH:
CO2 + 2NaOH -> Na2CO3 + H2O (2)
According to equation (2), 1 mole of CO2 reacts with 2 moles of NaOH to produce 1 mole of Na2CO3.
Since the volume is directly proportional to the number of moles of gas, we can represent this as:
CO2 + 2NaOH -> Na2CO3 + H2O
1 volume + 2 volumes -> 1 volume
Therefore, the 30 mL of the original mixture would react with twice the volume of NaOH, which is 30 mL * 2 = 60 mL.
Thus, the residual volume after treating 30 mL of the original mixture with aqueous NaOH would be (30 - 60) mL = -30 mL. However, negative volumes are not physically meaningful, so we can say that the residual volume would be 0 mL.
To solve this problem, we need to understand the chemistry behind the reactions involved.
When a mixture of CO (carbon monoxide) and CO2 (carbon dioxide) is sparked with oxygen (O2), combustion reactions occur, and the gases get converted into different compounds. Without specific information, we will assume that the combustion of CO and CO2 leads to the formation of carbon dioxide (CO2) only. Therefore, the balanced equation for the combustion reaction is:
2CO + O2 -> 2CO2
Now, let's analyze the given information step by step:
1. The original mixture has a volume of 20 mL, consisting of CO and CO2.
2. "x" mL of oxygen is added to the mixture and sparked.
3. After the explosion, the volume is reduced to (16+x) mL.
To find the residual volume after treating 30 mL of the original mixture with aqueous NaOH, we need to consider the reactions that occur:
1. CO2 reacts with NaOH to form sodium carbonate (Na2CO3) and water (H2O).
CO2 + 2NaOH -> Na2CO3 + H2O
2. CO does not react with NaOH; therefore, it remains unchanged.
Since we are using 30 mL of the original mixture, we assume the proportions of CO and CO2 are in a 1:1 ratio (10 mL each).
Now, let's calculate the volume of gas that remains after the reaction:
1. CO2 reacts with NaOH: 10 mL of CO2 will react according to the balanced equation.
CO2 + 2NaOH -> Na2CO3 + H2O
Since the stoichiometric ratio is 1:1 between CO2 and Na2CO3, the entire CO2 reacts to form Na2CO3.
2. Volume of CO that remains: Since CO does not react with NaOH, the entire 10 mL of CO will remain unchanged.
Therefore, the residual volume after treating 30 mL of the original mixture with NaOH is 10 mL.