Ethane (C2H6) burns with oxygen to produce carbon dioxide and water. A sample of ethane was burned completely and the water that formed had a mass of 1.61g. How much ethane, in moles and in grams, was in the sample?
Write the equation and balance it.
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Here is a stoichiometry problem I've posted. Just follow the steps in the example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the amount of ethane in the sample, we need to use the information about the water formed during the combustion reaction.
1. We first need to determine the molar mass of water (H2O). The molar mass of hydrogen (H) is 1.01 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. Therefore, the molar mass of water is:
(2 * 1.01 g/mol) + (1 * 16.00 g/mol) = 18.02 g/mol
2. Knowing the molar mass of water, we can calculate the number of moles of water formed using its mass and molar mass. The mass of water formed is given as 1.61 g. Therefore, the number of moles of water is:
Moles = Mass / Molar mass
Moles = 1.61 g / 18.02 g/mol ≈ 0.0894 mol (rounded to 4 decimal places)
3. In the balanced chemical equation for the combustion of ethane, we can see that for every mole of ethane burned, 3 moles of water are formed. Therefore, based on the moles of water formed, we can determine the moles of ethane in the sample:
Moles of ethane = Moles of water / 3
Moles of ethane = 0.0894 mol / 3 ≈ 0.0298 mol (rounded to 4 decimal places)
4. Finally, to find the mass of ethane in the sample, we can use the molar mass of ethane. The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Therefore, the molar mass of ethane is:
(2 * 12.01 g/mol) + (6 * 1.01 g/mol) = 30.07 g/mol
Mass of ethane = Moles of ethane * Molar mass of ethane
Mass of ethane = 0.0298 mol * 30.07 g/mol ≈ 0.893 g (rounded to 3 decimal places)
Therefore, the sample of ethane contained approximately 0.0298 moles or 0.893 grams of ethane.
To determine the amount of ethane in the sample, we can use the balanced chemical equation and the mass of water formed.
The balanced chemical equation for the combustion of ethane is:
C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O
From the equation, we can see that for every 3 moles of water produced, one mole of ethane is consumed. Therefore, the moles of ethane can be calculated as follows:
Moles of ethane = Moles of water / (3 moles of water / 1 mole of ethane)
To calculate the moles of ethane:
Moles of ethane = 1.61 g H2O / (18.015 g/mol H2O) / (3 mol H2O / 1 mol C2H6) = 0.0147 mol C2H6
So, there were 0.0147 moles of ethane in the sample.
To calculate the mass of ethane in grams, you can use the molar mass of ethane, which is:
Molar mass of C2H6 = (2 x Atomic mass of C) + (6 x Atomic mass of H) = (2 x 12.01 g/mol) + (6 x 1.01 g/mol) = 30.07 g/mol
Mass of ethane = Moles of ethane x Molar mass of ethane
Mass of ethane = 0.0147 mol x 30.07 g/mol = 0.44 g
Therefore, there were 0.0147 moles (or 0.44 grams) of ethane in the sample.