2C2H6(g) + 7O2 -> 4CO2 (g) + 6 H2o (g)
What quantity of reactant remains after ignition of a mixture that contains 0.40 moles of C2H6 mixed with 1.60 moles of O2? ( assume maximum reaction according to the equation above)
.20 mol of o2
To find out what quantity of reactant remains after ignition, we first need to determine the limiting reactant. The limiting reactant is the one that will be completely consumed first and will determine the amount of product formed.
Let's convert the moles of C2H6 and O2 into moles of CO2 using the stoichiometry of the balanced equation:
For 1 mole of C2H6, we produce 4 moles of CO2.
0.40 moles of C2H6 x (4 moles CO2 / 2 moles C2H6) = 0.80 moles of CO2.
For 1 mole of O2, we produce 0.40 moles of CO2.
1.60 moles of O2 x (4 moles CO2 / 7 moles O2) = 0.91 moles of CO2.
From the calculations, we can see that 0.80 moles of CO2 is produced from the C2H6 and 0.91 moles of CO2 is produced from the O2. Since we need 4 moles of CO2 for every 2 moles of C2H6, and there is an equal amount of O2 available for both reactants, the C2H6 is the limiting reactant.
To find out what quantity of C2H6 remains, we subtract the amount reacted from the initial amount:
0.40 moles of C2H6 - 0.40 moles of C2H6 reacted = 0 moles of C2H6 remaining.
Therefore, after ignition, there is no C2H6 remaining.
To find out the quantity of reactant that remains after the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed during the reaction and determines the maximum amount of product that can be formed.
In this case, we have 0.40 moles of C2H6 (ethane) and 1.60 moles of O2 (oxygen). We can compare the number of moles of each reactant to their stoichiometric coefficients in the balanced equation.
From the balanced equation:
2 moles of C2H6 react with 7 moles of O2 to produce 4 moles of CO2 and 6 moles of H2O.
To determine the limiting reactant, we need to calculate the moles of CO2 and H2O that can be formed from each reactant. We will use the moles of O2 as a basis for comparison.
Moles of CO2 from C2H6 = (0.40 moles C2H6) * (4 moles CO2 / 2 moles C2H6) = 0.80 moles CO2
Moles of H2O from C2H6 = (0.40 moles C2H6) * (6 moles H2O / 2 moles C2H6) = 1.20 moles H2O
Moles of CO2 from O2 = (1.60 moles O2) * (4 moles CO2 / 7 moles O2) = 0.92 moles CO2
Moles of H2O from O2 = (1.60 moles O2) * (6 moles H2O / 7 moles O2) = 1.37 moles H2O
Comparing the moles of CO2 and H2O produced by each reactant, we can see that the moles of CO2 from C2H6 (0.80 moles) are less than the moles of CO2 from O2 (0.92 moles). Similarly, the moles of H2O from C2H6 (1.20 moles) are less than the moles of H2O from O2 (1.37 moles).
The limiting reactant is the one that produces the lesser amount of the product. In this case, C2H6 is the limiting reactant because it produces fewer moles of both CO2 and H2O.
Now that we know C2H6 is the limiting reactant, we can calculate the quantity of C2H6 that is completely consumed during the reaction. Since the stoichiometric coefficient of C2H6 is 2 in the balanced equation, the moles of C2H6 consumed is given by:
Moles of C2H6 consumed = (0.40 moles C2H6) - (moles of CO2 produced * (2 moles C2H6 / 4 moles CO2)) = 0.40 - (0.80 * (2/4)) = 0.40 - 0.40 = 0 moles
Therefore, after the reaction, 0.40 moles of C2H6 remain unreacted.
This is a limiting reagent problem. Here is an example of such a problem. Just follow the steps to find the limiting reagent, then work backwards to find how much of the non-limiting reagent is used. Finally, subtract amount used from initial amount to determine amount remaining.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html