Question

A canister contains butane at a pressure of 760 torr and a temprature of 25C. All of contents of the canister are allowed to react with excess oxygen to produce carbon dioxide and water.

2 (C4H10) + 13O2 -------> 8CO2+ 10(H2O)

Equation is balanced and butane is completely consumed in the reaction, and 46.5g grams of CO2 are produced. What is the volume, in liters, of the canister that contained the butane??

Answer is 6.46L but can anyone please show work for this question?? I just can't get the right answer or near even near it. Please teach me how to approach this problem. Thank You!

Answer 1

pressure = 760 torr = 1 atm

temprature = 25C = 298 K

mass of CO2 produced = 46.5 g

moles of CO2 produced = 46.5 / 44

                                       = 1.0568 moles

2 (C4H10) + 13O2 ------------------> 8CO2+ 10(H2O)

here

2 moles of (C4H10) gives   ------------------>    8 moles of CO2

? moles of (C4H10) gives   -------------------> 1.0568 moles of CO2

moles of (C4H10) = 1.0568 x 2 / 8

                             = 0.2642

from the ideal gas equation

PV = nRT

1 x V = 0.2642 x 0.0821 x 298

V = 6.46 L

the volume, in liters, of the canister that contained the butane is 6.46 L