Write the balanced chemical equation for the combustion of ethane, C2H6, and answer these questions: (Use the lowest possible coefficients. Omit states of matter.) How many molecules of oxygen would combine with 22 molecules of ethane in this reaction? Molecules of oxygen If 13.2 mol of oxygen is consumed in the reaction, how many moles of water are produced? Moles of water How many grams of ethane are burned if 4.0 * 10^22 molecules of CO2 are produced? Ethane
The Correct Answer and Explanation is:
Balanced Chemical Equation for the Combustion of Ethane:
The combustion of ethane involves a reaction with oxygen to produce carbon dioxide and water. The general equation is:C2H6+O2→CO2+H2OC_2H_6 + O_2 \rightarrow CO_2 + H_2OC2H6+O2→CO2+H2O
To balance the equation, follow these steps:
- Balance the carbon atoms:
- There are 2 carbon atoms in ethane (C2H6), so you need 2 molecules of CO2 to balance the carbon atoms.
- Balance the hydrogen atoms:
- There are 6 hydrogen atoms in ethane, so you need 3 molecules of H2O to balance the hydrogen atoms (since each H2O has 2 hydrogen atoms).
- Balance the oxygen atoms:
- On the right side of the equation, you now have 2 molecules of CO2, each containing 2 oxygen atoms (total 4 oxygens), and 3 molecules of H2O, each containing 1 oxygen atom (total 3 oxygens). Therefore, the total oxygen atoms required is 7.
- To balance this with oxygen on the left, place a coefficient of 7/2 in front of O2.
- Eliminate fractional coefficients by multiplying through by 2:
- This gives the final balanced equation:
2C2H6+7O2→4CO2+6H2O2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O2C2H6+7O2→4CO2+6H2O
Answering the Questions:
- How many molecules of oxygen would combine with 22 molecules of ethane in this reaction?
- According to the balanced equation, 2 molecules of C2H6 react with 7 molecules of O2.
- Therefore, the ratio of oxygen to ethane is 7:2.
- For 22 molecules of ethane:
Molecules of O2=(72)×22=77 molecules of O2.\text{Molecules of O2} = \left(\frac{7}{2}\right) \times 22 = 77 \text{ molecules of O2}.Molecules of O2=(27)×22=77 molecules of O2.
- If 13.2 mol of oxygen is consumed in the reaction, how many moles of water are produced?
- From the balanced equation, 7 moles of O2 produce 6 moles of H2O.
- So, if 13.2 moles of oxygen are consumed, the amount of water produced is:
Moles of H2O=(67)×13.2=11.3 moles of H2O.\text{Moles of H2O} = \left(\frac{6}{7}\right) \times 13.2 = 11.3 \text{ moles of H2O}.Moles of H2O=(76)×13.2=11.3 moles of H2O.
- How many grams of ethane are burned if 4.0 × 10^22 molecules of CO2 are produced?
- From the balanced equation, 2 molecules of ethane produce 4 molecules of CO2.
- Thus, for 4.0 × 10^22 molecules of CO2:
Molecules of C2H6=(24)×4.0×1022=2.0×1022 molecules of C2H6.\text{Molecules of C2H6} = \left(\frac{2}{4}\right) \times 4.0 \times 10^{22} = 2.0 \times 10^{22} \text{ molecules of C2H6}.Molecules of C2H6=(42)×4.0×1022=2.0×1022 molecules of C2H6.
- The molar mass of ethane (C2H6) is approximately 30.07 g/mol.
- To find the mass of ethane burned, first calculate the number of moles of ethane:
Moles of C2H6=2.0×10226.022×1023=3.32×10−2 moles of C2H6.\text{Moles of C2H6} = \frac{2.0 \times 10^{22}}{6.022 \times 10^{23}} = 3.32 \times 10^{-2} \text{ moles of C2H6}.Moles of C2H6=6.022×10232.0×1022=3.32×10−2 moles of C2H6.
- Finally, multiply by the molar mass to get the mass:
Mass of C2H6=3.32×10−2×30.07≈1.0 grams of C2H6.\text{Mass of C2H6} = 3.32 \times 10^{-2} \times 30.07 \approx 1.0 \text{ grams of C2H6}.Mass of C2H6=3.32×10−2×30.07≈1.0 grams of C2H6.
Explanation:
The balanced chemical equation tells us that ethane reacts with oxygen in a 2:7 ratio. This helps us calculate the amounts of reactants and products in terms of molecules, moles, and mass. The molar relationships in the equation are crucial for converting between units like moles, molecules, and grams. In each case, stoichiometry allows us to determine how much oxygen, water, or ethane is involved based on the quantities provided.
