ethane,C2H6 burns in oxygen.
A. what are the products of the reaction?
B. write the balanced equation for the reaction.
C. what mass of O2, in grams is required for complete combustion of 13.6 of ethane?
D. what is the total mass of products expected from the combustion of 13.6 g of ethane?
what's the answer
A. To determine the products of the reaction between ethane (C2H6) and oxygen (O2), we need to consider the combustion reaction. Combustion is a type of chemical reaction where a compound reacts with oxygen to form carbon dioxide (CO2) and water (H2O). Therefore, the products of the reaction between ethane and oxygen are carbon dioxide and water.
B. To write the balanced equation for the combustion of ethane, we need to ensure that the number of atoms on both sides of the equation is balanced. The molecular formula for ethane is C2H6, and for oxygen gas, it is O2. The balanced equation for the reaction can be written as:
2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O
C. To determine the mass of O2 required for the complete combustion of 13.6 g of ethane, we need to use stoichiometry. From the balanced equation above, we can see that 2 moles of ethane react with 7 moles of oxygen. We can calculate the molar mass of ethane (C2H6) and the molar mass of oxygen (O2) to convert grams to moles. The molar mass of ethane is calculated by adding the atomic masses of carbon (12.01 g/mol) multiplied by 2, and the atomic mass of hydrogen (1.01 g/mol) multiplied by 6, giving a total molar mass of 30.07 g/mol. The molar mass of oxygen gas (O2) is 32.00 g/mol.
Using the equation:
(13.6 g ethane) x (1 mol ethane / 30.07 g ethane) x (7 mol oxygen / 2 mol ethane) x (32.00 g oxygen / 1 mol oxygen) = mass of oxygen needed
By substituting the values, we find that approximately 77.0 g of oxygen is required for the complete combustion of 13.6 g of ethane.
D. To calculate the total mass of products expected from the combustion of 13.6 g of ethane, we need to consider the stoichiometry of the balanced equation. From the balanced equation, we can see that for every 2 moles of ethane, we obtain 4 moles of carbon dioxide and 6 moles of water.
Using the equation:
(13.6 g ethane) x (1 mol ethane / 30.07 g ethane) x (4 mol carbon dioxide / 2 mol ethane) x (44.01 g carbon dioxide / 1 mol carbon dioxide) + (13.6 g ethane) x (1 mol ethane / 30.07 g ethane) x (6 mol water / 2 mol ethane) x (18.02 g water / 1 mol water) = mass of products
Calculating the values, we find that the total mass of products expected from the combustion of 13.6 g of ethane is approximately 45.3 g.