Ethane (C2H6) reacts with oxygen to produce carbon dioxide and water. Assuming there is an excess of oxygen, calculate the mass of CO2 and H2O produced from 1.25 g of ethane
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Ethane C2H6 reacts with oxygen gas O2 to produce carbon dioxide and water
To calculate the mass of CO2 and H2O produced from 1.25 g of ethane (C2H6), we will first start by writing and balancing the chemical equation for the reaction:
C2H6 + O2 → CO2 + H2O
The balanced equation indicates that 1 mole of ethane reacts with 3.5 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water.
Now, we can calculate the molar mass of ethane (C2H6):
C: 2 atoms x 12.01 g/mol = 24.02 g/mol
H: 6 atoms x 1.01 g/mol = 6.06 g/mol
Therefore, the molar mass of ethane (C2H6) is 24.02 g/mol + 6.06 g/mol = 30.08 g/mol.
Next, we need to determine the number of moles of ethane we have:
1.25 g of ethane x (1 mol ethane / 30.08 g ethane) = 0.0415 mol ethane
Now, we can calculate the number of moles of CO2 produced. From the balanced equation, we know that 1 mole of ethane produces 2 moles of CO2. Therefore:
0.0415 mol ethane x (2 mol CO2 / 1 mol ethane) = 0.083 mol CO2
To calculate the mass of CO2 produced, we can use the molar mass of CO2, which is 44.01 g/mol:
0.083 mol CO2 x 44.01 g CO2/mol = 3.65 g CO2
Finally, we can calculate the number of moles of H2O produced. From the balanced equation, we know that 1 mole of ethane produces 3 moles of H2O. Therefore:
0.0415 mol ethane x (3 mol H2O / 1 mol ethane) = 0.124 mol H2O
To calculate the mass of H2O produced, we can use the molar mass of H2O, which is 18.02 g/mol:
0.124 mol H2O x 18.02 g H2O/mol = 2.24 g H2O
Therefore, from 1.25 g of ethane, we would produce approximately 3.65 g of CO2 and 2.24 g of H2O.
To calculate the mass of CO2 and H2O produced from a given mass of ethane, we need to use the balanced chemical equation for the reaction between ethane (C2H6) and oxygen (O2).
The balanced equation for the reaction is as follows:
C2H6 + 7/2 O2 → 2 CO2 + 3 H2O
From the balanced equation, we can see that for every 2 moles of CO2, there are 1 mole of C2H6, and for every 3 moles of H2O, there are 1 mole of C2H6.
First, we need to determine the number of moles of ethane given the mass of 1.25 g. To do this, we use the molar mass of ethane, which can be calculated by adding up the atomic masses of carbon (C) and hydrogen (H) in the molecular formula C2H6.
Molar mass of C2H6 = (2 * molar mass of C) + (6 * molar mass of H)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol
Number of moles of ethane = Given mass / Molar mass
= 1.25 g / 30.08 g/mol
≈ 0.0416 mol (rounded to four decimal places)
Now that we have the number of moles of ethane, we can calculate the moles of CO2 and H2O produced using the mole ratios from the balanced equation.
For every 1 mole of C2H6, we get 2 moles of CO2. Therefore, the number of moles of CO2 produced is:
Number of moles of CO2 = Number of moles of C2H6 * (2 moles of CO2 / 1 mole of C2H6)
= 0.0416 mol * (2 moles of CO2 / 1 mole of C2H6)
= 0.0832 mol
For every 1 mole of C2H6, we get 3 moles of H2O. Therefore, the number of moles of H2O produced is:
Number of moles of H2O = Number of moles of C2H6 * (3 moles of H2O / 1 mole of C2H6)
= 0.0416 mol * (3 moles of H2O / 1 mole of C2H6)
= 0.1248 mol
To calculate the mass of CO2 and H2O produced, we need to multiply the respective number of moles by their molar masses.
Mass of CO2 = Number of moles of CO2 * Molar mass of CO2
= 0.0832 mol * 44.01 g/mol (molar mass of CO2)
= 3.67 g (rounded to two decimal places)
Mass of H2O = Number of moles of H2O * Molar mass of H2O
= 0.1248 mol * 18.02 g/mol (molar mass of H2O)
= 2.25 g (rounded to two decimal places)
Therefore, from 1.25 g of ethane, approximately 3.67 g of CO2 and 2.25 g of H2O will be produced assuming there is an excess of oxygen.