104 ml of ethene, C2H4 is burned in 408 ml of oxygen, producing carbon dioxide and some liquid water. Some oxygen remain unreacted. Calculate the volume of carbon dioxide produced and the volume of oxygen remaining
balance the reaction equation.
C2H4+3O2>>>2CO2 + 2H2O
if one assumes temp and pressure remain constant, then the law of volumes indicates one gets two volumes of carbon dioxide for each volume of ethene.
So ...208ml
volume of O2 remaining, you used 3*104ml, so subtract that from 408ml.
All this is very wild speculation of constant temp,pressure.
To calculate the volume of carbon dioxide produced and the volume of oxygen remaining, we need to balance the chemical equation and use the concept of stoichiometry.
First, let's write the balanced chemical equation for the combustion of ethene:
C2H4 + O2 -> CO2 + H2O
From the balanced equation, we can see that one mole of ethene reacts with one mole of oxygen to produce one mole of carbon dioxide.
Next, we need to determine the number of moles of ethene and oxygen that we have based on the given volumes and the ideal gas law. The ideal gas law equation is:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
Volume of ethene (C2H4) = 104 ml
Volume of oxygen (O2) = 408 ml
Now, we need to convert the volumes from milliliters to liters. Since 1 liter is equal to 1000 milliliters, we divide each volume by 1000:
Volume of ethene (C2H4) = 104 ml / 1000 = 0.104 L
Volume of oxygen (O2) = 408 ml / 1000 = 0.408 L
Now we can use the ideal gas law to calculate the number of moles of ethene and oxygen. Assuming the reaction is done at standard temperature and pressure (STP), where T = 273.15K and P = 1 atm, we can use the ideal gas constant R = 0.0821 L.atm/mol.K:
n(ethene) = (P x V) / (R x T) = (1 atm x 0.104 L) / (0.0821 L.atm/mol.K x 273.15K)
Now, we need to calculate the number of moles of oxygen:
n(oxygen) = (P x V) / (R x T) = (1 atm x 0.408 L) / (0.0821 L.atm/mol.K x 273.15K)
Now that we have the number of moles of ethene and oxygen, we can determine the volume of carbon dioxide produced and the volume of oxygen remaining based on the stoichiometry of the balanced equation.
From the balanced equation, we know that the molar ratio between ethene and carbon dioxide is 1:1. Therefore, the number of moles of carbon dioxide produced will be equal to the number of moles of ethene consumed.
Volume of carbon dioxide produced = n(ethene)
To calculate the moles of oxygen remaining, we need to subtract the number of moles of oxygen reacted from the initial number of moles of oxygen.
Moles of oxygen remaining = n(oxygen initial) - n(oxygen reacted)
Finally, we can calculate the volume of oxygen remaining using the ideal gas law:
Volume of oxygen remaining = (moles of oxygen remaining x R x T) / P
Now that we have the calculations explained, you can substitute the values and perform the necessary calculations.