calculate when 7.14 of butane reacts with excess oxygen
To calculate when 7.14 grams of butane (C4H10) reacts with excess oxygen, we need to use stoichiometry to determine the balanced equation for the reaction.
The balanced equation for the combustion of butane is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
This balanced equation tells us that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
To find the amount of oxygen needed to react with 7.14 grams of butane, we'll need to convert the given mass of butane to moles.
The molar mass of butane (C4H10) can be calculated as follows:
C (12.01 g/mol) * 4 + H (1.01 g/mol) * 10 = 58.12 g/mol
Now, we can calculate the number of moles of butane using the given mass:
7.14 g / 58.12 g/mol ≈ 0.123 moles of butane
Since it reacts with excess oxygen, all the butane will be consumed in the reaction. Therefore, we can infer that 0.123 moles of butane will react.
According to the balanced equation, 2 moles of butane react with 13 moles of oxygen. Therefore, we can calculate the amount of oxygen needed:
(0.123 moles of butane) * (13 moles of oxygen / 2 moles of butane) ≈ 0.7995 moles of oxygen
Now, to calculate the time it takes for this reaction to occur, we would need more information. The rate of reaction depends on various factors such as temperature, pressure, and the presence of a catalyst. If this information is given, it can be used to determine the reaction time based on the rate constant and reaction kinetics. However, without this additional information, it is not possible to provide a specific time for the reaction to take place.