Balanced eqn: 2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(l) You feed the piston wit

Question

Balanced eqn: 2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(l)

You feed the piston with 3.0 mol of C8H18 and 45.0 mol of O2 and have them react.

What would I do to find limiting reactant?

Also, how would you determine the work done (in joules) when a sample of gas (product of above reaction) extends the position of a test piston from a volume of 552 mL to 891 mL at constant temperature against a constant 1.25 atm pressure (1 L*atm = 101.3 J)

Best answer for showing me how to come up with the answers, not just handing me the answers.

Explanation / Answer

Balanced eqn: 2C8H18(l) + 25O2(g) -> 16CO2(g) + 18H2O(l)

2 moles of C8H18 reacts with 25 moles of O2.

limiting reagent is one which is consumed first in the reaction and determines the moles of the products formed. Here, 2 moles of C8H18 reacts with 25 moles of O2. Thus, one moleof C8H18 react with 12.5 moles of O2. And hence, 3 moles of C8H18 will react with 37.5 moles of O2. Thus, C8H18 will be consumed first and will act as a limiting reagent.

Work = -P(deltaV) = - 1.25 x (.891-.552) latm = -42.9259 J

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