c2h4(g) + 3O2(g) -> 2 CO2 (g) + 2 H2O (g)
What volume of oxygen will react with 18 ml of ehtylene, c2h4, assuming that the gases are present at the same temperature and pressure?
When using gases only in an equation, we don't need to convert to moles but we can use the volume and they is proportional to the moles. Your post doesn't specify but I assume there is enough oxygen to use all of the ethylene.
18 mL C2H2 x (3 moles O2/1 mole C2H4) = ??
To find the volume of oxygen that will react with 18 ml of ethylene, we first need to determine the stoichiometric ratio between ethylene (C2H4) and oxygen (O2) in the balanced chemical equation.
From the balanced equation:
C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(g)
We can see that the stoichiometric ratio between ethylene and oxygen is 1:3. This means that for every 1 mole of ethylene, we need 3 moles of oxygen to react completely.
To find the volume of oxygen, we can use the following equation:
Volume = (Number of moles * Molar volume) / Molar ratio
First, we need to convert the volume of ethylene (C2H4) into moles. To do this, we need the molar mass of ethylene, which is:
(2 * Atomic mass of carbon) + (4 * Atomic mass of hydrogen)
= (2 * 12.01 g/mol) + (4 * 1.01 g/mol)
= 24.02 g/mol + 4.04 g/mol
= 28.06 g/mol
Now, we can calculate the number of moles of ethylene:
Number of moles = Volume (in liters) / Molar volume
= 18 ml / 1000 ml/L
= 0.018 L / 1 L
= 0.018 moles
Next, we can use the stoichiometric ratio between ethylene and oxygen to calculate the number of moles of oxygen:
Number of moles of oxygen = Number of moles of ethylene * Molar ratio
= 0.018 moles * 3
= 0.054 moles
Finally, we can calculate the volume of oxygen using the molar volume at the same temperature and pressure:
Volume of oxygen = (Number of moles of oxygen * Molar volume) / Molar ratio
= (0.054 moles * 22.4 L/mol) / 3
= 1.296 L / 3
= 0.432 L
Therefore, the volume of oxygen that will react with 18 ml of ethylene is approximately 0.432 liters.