If l burn 6grams in plenty of air, what volume of oxygen will be used up
To find the volume of oxygen consumed when burning 6 grams of a substance, we need to know the chemical formula of the substance. The balanced chemical equation of the combustion reaction can provide the necessary information.
However, if we assume that the substance you are burning is a hydrocarbon and completely reacts with oxygen to produce carbon dioxide (CO2) and water (H2O), we can estimate the volume of oxygen consumed using the stoichiometry of the reaction.
Based on the balanced chemical equation:
C(s) + O2(g) -> CO2(g) + H2O(g)
The stoichiometry of the reaction shows that 1 mole of carbon (12 grams) reacts with 1 mole of oxygen (32 grams) to produce 1 mole of carbon dioxide and water.
Therefore, the molar ratio between carbon and oxygen is 1:1.
Since you are burning 6 grams of the substance, and it contains only carbon, we can calculate the amount of oxygen needed.
6 grams of carbon x (1 mole of oxygen / 12 grams of carbon) = 0.5 moles of oxygen
Now, to find the volume of oxygen consumed, we need to know the conditions (temperature and pressure) under which the reaction is taking place. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. So, we can convert the moles of oxygen to liters by using this value:
0.5 moles of oxygen x 22.4 liters/mole = 11.2 liters of oxygen
Therefore, approximately 11.2 liters of oxygen gas would be consumed when burning 6 grams of the given substance, assuming it is a hydrocarbon and that the reaction takes place at standard temperature and pressure (STP).
To determine the volume of oxygen that will be used up when burning 6 grams of a substance in plenty of air, we need to consider the stoichiometry of the chemical reaction involved.
First, we need to know the balanced chemical equation for the combustion reaction. Let's assume we are burning a hydrocarbon such as butane (C4H10):
C4H10 + 13O2 → 8CO2 + 10H2O
By examining the balanced equation, we can see that it takes 13 moles of oxygen (O2) to completely burn 1 mole of butane (C4H10).
Now, let's calculate the molar mass of oxygen:
Molar mass of O2 = 2 * molar mass of oxygen
= 2 * 16 g/mol
= 32 g/mol
From the molar mass, we know that 1 mole of oxygen weighs 32 grams.
To find the moles of oxygen required to burn 6 grams of butane, we can use the conversion factor:
6 g butane * (1 mole butane / molar mass of butane)
* (13 moles of O2 / 1 mole of butane)
* (1 mole of oxygen / 13 moles of O2)
* (molar mass of oxygen / 1 mole of oxygen)
Substituting the values, we get:
6 g * (1 mol / 58.12 g) * (13 mol / 1 mol) * (1 mol / 13 mol) * (32 g / 1 mol)
= 32 g
Therefore, 32 grams of oxygen will be consumed during the combustion of 6 grams of butane.
As for determining the volume of oxygen used, we need to know the standard molar volume at a given temperature and pressure. At standard temperature and pressure (STP), which is 0 degrees Celsius and 1 atmosphere of pressure, 1 mole of any ideal gas occupies 22.4 liters.
So, if we assume that the combustion occurs at STP, we can calculate the volume of oxygen used:
Volume of oxygen = (moles of oxygen) * (22.4 L/mole)
For our case, the moles of oxygen is 32 g / molar mass of oxygen.
Plugging in the values, we can determine the volume of oxygen used in liters.
air is about 21% oxygen and 78% nitrogen, so one mole of air weighs about .21*32 + .78*28 = 28.56 grams. That means you have
0.21 moles of air.
one mole of a gas occupies 22.4L, so now you can see how much of that is O2.