Answer:
To balance this equation, we need to know the coefficients for each reactant and product. The balanced equation for the reaction is:
C7H17 + 11O2 → 7CO2 + 8H2O
This equation indicates that one molecule of heptane reacts with 11 molecules of oxygen gas to produce seven molecules of carbon dioxide and eight molecules of water.
The stoichiometric coefficients in the balanced equation tell us the relative amounts of each reactant and product in the reaction. For example, one molecule of heptane reacts with 11 molecules of oxygen gas to produce seven molecules of carbon dioxide and eight molecules of water. Therefore, if we have 1 mole of heptane, we need 11 moles of oxygen gas to completely react with it, and we will produce 7 moles of carbon dioxide and 8 moles of water.
To determine the amount of each product that will be produced when a given amount of reactants is used, we need to use stoichiometry. We can convert the mass of heptane and oxygen gas to moles, and then use the stoichiometric coefficients in the balanced equation to calculate the amount of carbon dioxide and water that should be produced.
For example, if we have 10 grams of heptane and 50 grams of oxygen gas, we can calculate the amount of each product that should be produced as follows:
1. Calculate the number of moles of each reactant:
moles of heptane = 10 g / 100.21 g/mol = 0.0999 mol
moles of oxygen gas = 50 g / 31.9988 g/mol = 1.562 mol
2. Determine the limiting reactant:
Using the stoichiometric coefficients in the balanced equation, we can see that 1 mole of heptane reacts with 11 moles of oxygen gas. Therefore, the limiting reactant is heptane, because we only have 0.0999 moles of it, whereas we have 1.562 moles of oxygen gas.
3. Calculate the theoretical yield of each product:
moles of CO2 = moles of heptane × (7 moles of CO2 / 1 mole of C7H17) = 0.6993 mol
moles of H2O = moles of heptane × (8 moles of H2O / 1 mole of C7H17) = 0.7992 mol
mass of CO2 = moles of CO2 × molar mass of CO2 = 0.6993 mol × 44.01 g/mol = 30.77 g
mass of H2O = moles of H2O × molar mass of H2O = 0.7992 mol × 18.015 g/mol = 14.39 g
Therefore, if we have 10 grams of heptane and 50 grams of oxygen gas, the theoretical yield of carbon dioxide is 30.77 grams and the theoretical yield of water is 14.39 grams. Note that the actual yield may be different, depending on the conditions of the reaction and the efficiency of the reaction.