In the reaction O2+2NO->2NO2 what is the ratio of oxygen reacted to nitrogen dioxide produced? Nitrogen monoxide to nitrogen dioxide?
We had to calculate the concentration of each, for which I have O2=-1.01x10^-5, NO=-2.03x10^-5 and NO2=2.04x10^-5.
I'm not sure how to do it but what I got was 1O2:2NO2 and 1NO:1NO2.
We also have to compare it to the balanced equation and I don't understand what to do for that part.
ratio? O/NO=1/2
No/NO2=2/2=1/1
To determine the ratios of oxygen reacted to nitrogen dioxide produced and nitrogen monoxide to nitrogen dioxide, you can use the coefficients in the balanced equation.
First, let's write the balanced equation for the reaction:
O2 + 2NO -> 2NO2
From the equation, we can see that one molecule of O2 reacts to produce two molecules of NO2.
Therefore, the ratio of oxygen reacted to nitrogen dioxide produced is 1:2 (1 O2 : 2 NO2).
Similarly, from the equation, we can see that two molecules of NO react to produce two molecules of NO2.
Therefore, the ratio of nitrogen monoxide to nitrogen dioxide is 2:2 (2 NO : 2 NO2), which simplifies to 1:1 (1 NO : 1 NO2).
To compare these ratios with the concentrations you provided, you'll need to convert the concentrations to moles.
For O2:
Concentration (O2) = -1.01x10^-5 M
1 mole of gas occupies 22.4 liters at standard temperature and pressure (STP). Assuming this condition, you can convert the concentration to moles by multiplying it by the volume of the gas in liters. Furthermore, you need the moles of O2 reacting, i.e., multiplying by the coefficient in the balanced equation.
Moles of O2 = (-1.01x10^-5 M) * (22.4 L) * (1 mole O2 / 1 L) = -0.0002264 moles O2
For NO:
Concentration (NO) = -2.03x10^-5 M
Moles of NO = (-2.03x10^-5 M) * (22.4 L) * (2 moles NO / 1 L) = -0.0009104 moles NO
For NO2:
Concentration (NO2) = 2.04x10^-5 M
Moles of NO2 = (2.04x10^-5 M) * (22.4 L) * (2 moles NO2 / 1 L) = 0.000911 moles NO2
Comparing the moles:
O2 : NO2 = -0.0002264 moles O2 : 0.000911 moles NO2 (approximately 1 : 4)
NO : NO2 = -0.0009104 moles NO : 0.000911 moles NO2 (approximately 1 : 1)
Therefore, the ratios based on the concentrations agree with the ratios predicted by the balanced equation.
To determine the ratio of oxygen reacted to nitrogen dioxide produced in the reaction O2 + 2NO -> 2NO2, we can compare the coefficients in the balanced equation.
In the balanced equation, the coefficient in front of O2 is 1, indicating that 1 mole of O2 reacts. The coefficient in front of NO2 is also 1, indicating that 1 mole of NO2 is produced. Therefore, the ratio of oxygen reacted to nitrogen dioxide produced is 1:1.
Similarly, to determine the ratio of nitrogen monoxide (NO) to nitrogen dioxide (NO2), we can again look at the coefficients. In the balanced equation, the coefficient in front of NO is 2, indicating that 2 moles of NO react. The coefficient in front of NO2 is also 2, indicating that 2 moles of NO2 are produced. Therefore, the ratio of NO to NO2 is 2:2, which simplifies to 1:1.
Now, let's compare the calculated concentrations you provided (O2 = -1.01x10^-5, NO = -2.03x10^-5, NO2 = 2.04x^-5) with the ratios determined from the balanced equation.
From the concentrations you provided, the ratio of O2 to NO2 can be calculated by dividing the concentration of O2 by the concentration of NO2: (-1.01x10^-5) / (2.04x10^-5) ≈ 0.495, which is approximately 1:2 (rounded to the nearest whole number).
For the ratio of NO to NO2, divide the concentration of NO by the concentration of NO2: (-2.03x10^-5) / (2.04x10^-5) ≈ 0.995, which is approximately 1:1 (rounded to the nearest whole number), matching the ratio from the balanced equation.
Comparing these ratios with the ones from the balanced equation allows us to see if the experimental data aligns with the stoichiometry indicated by the chemical equation. In this case, the experimental ratios do not perfectly match the ratios indicated by the balanced equation, but they are close.