How many moles of butane on complete combustion gives 18 g of water?
To determine the number of moles of butane that gives 18 g of water, we first need to set up the balanced chemical equation for the complete combustion of butane:
C4H10 + 13/2 O2 -> 4 CO2 + 5 H2O
From the equation, we can see that for every 13/2 moles of oxygen (O2), we obtain 5 moles of water (H2O). Therefore, the molar ratio between water and oxygen is 5:13/2, which simplifies to 10:13.
Now, we can calculate the number of moles of water based on the given mass and the molar mass of water:
Mass of water = 18 g
Molar mass of water (H2O) = 18 g/mol
Number of moles of water = Mass of water / Molar mass of water
= 18 g / 18 g/mol
= 1 mole
Since there is a 10:13 molar ratio between water and oxygen, the number of moles of oxygen required would be:
Number of moles of oxygen = Number of moles of water * (13/10)
= 1 mole * (13/10)
= 1.3 moles
Therefore, the complete combustion of butane that gives 18 g of water would require 1.3 moles of oxygen.
To find the number of moles of butane that will give 18 g of water upon complete combustion, we need to use the balanced equation for the combustion of butane.
The balanced equation is:
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O
From the equation, we can see that for every 5 moles of water produced, we need 1 mole of butane. Now, we can calculate the number of moles of butane.
Step 1: Calculate the molar mass of water.
The molar mass of H2O is:
(2 * molar mass of H) + (1 * molar mass of O)
= (2 * 1.008 g/mol) + (1 * 16.00 g/mol)
= 18.02 g/mol
Step 2: Convert the given mass of water to moles.
Moles of water = Given mass / molar mass
= 18 g / 18.02 g/mol
= 0.9994 mol (approximately)
Step 3: Determine the moles of butane.
Since we know that for every 5 moles of water, we have 1 mole of butane, we can set up a ratio:
1 mole of butane / 5 moles of water = Moles of butane / Moles of water
Rearranging the equation, we get:
Moles of butane = (1 mole of butane / 5 moles of water) * Moles of water
= (1/5) * 0.9994 mol
= 0.1999 mol (approximately)
Therefore, approximately 0.1999 moles of butane will give 18 g of water upon complete combustion.