Please show me how to do this. The balanced reaction for the combustion of ethan

Question

Please show me how to do this.

The balanced reaction for the combustion of ethane is shown in the table below You allow 3 2 mol of ethane (C_2H_6) to react with 4 4 mol oxygen (O_2) Complete the following ICF table to indicate the amounts of all chemical species for the initial change, and final conditions Drag the appropriate amounts to their respective targets. When solving problems involving stoichiometric coefficients the first step is to make sure you have a balanced chemical equation Then, you determine the limiting reagent by using the coefficients from the balanced equation You can keep track of the amounts of all reactant and products before and after reaction using an ICF table Completing the ICF table will also allow you to determine the limiting reagent, and the amount of product formed assumes that the reaction runs to completion with 100% yield Parts A and C explore these steps in more detail Let us consider another reaction. Ammonia and oxygen react to form nitrogen monoxide and water Construct your own balanced equation to determine the amount of NO and H_2O that would form when 3.37 mol NH_3 and 5 44 mol O_2 react Express the amounts in moles to two decimal places separated by a comma.

Explanation / Answer

Following is the ICE table for changes observed for combustion of ethane:

Explanation:

According to balanced equation 7 mol of oxygen combusts 2 mol of C2H6.

Hence 4.4 mol of oxygen combusts = 4.4 x 2 / 7 = 1.257 mol C2H6 which is approx. equal to 1.3 mol in the given options.

2 mol of C2H6 gives 4 mol of CO2.

1.3 mol of C2H6 gives 2.6 mol of CO2 which is close to 2.5 mol of CO2 in in the given options.

2 mol of C2H6 gives 6 mol of H2O.

1.3 mol of C2H6 gives 3.9 mol of H2O which is close to 3.8 mol of H2O in in the given options.

Finally the remaining amount of C2H6 can be obtained by subtracting the amount of C2H6 that underwent combustion from the initial amount of C2H6

Hence 3.2 mol of C2H6 - 1.3 mol of C2H6 subjected to combustion = 1.9 mol C2H6 remaining. Amount of oxygen left is 0 mol.   

2 C2H6 (g) 7 O2 (g) ……..> 4 CO2 (g) 6 H2O (g) Initial 3.2 mol 4.4 mol 0.0 mol 0.0 mol Change 1.3 mol 4.4 mol 2.5 mol 3.8 mol Final 1.9 mol 0.0 mol 2.5 mol 3.8 mol

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