Octane and Carbon dioxide

Question

a)   How many grams of CO₂ would be produced if 12 grams of octane reacts with an excess of oxygen? 

balanced 

2C₈H₁₈ + 25O_{2 -->} 16CO₂ + 18H₂O

 

 

b) What volume will the CO₂ gas collected in a) occupy at a temperature of 330K and a pressure of 0.533 atm? 

Answer 1

the first part of this question is just reaction stoichiometry, and since you're given that O2 is in excess, you dont need to use that.

 

12g C8H18 * (1 mol/114.26g C8H18) * (16 mol CO2/2mol C8H18)* (44.01 g CO2)=

 

given amount of reactant * 1/(molar mass) * (molar equivalent from balance equation) * (molar mass of CO2)

 

 

once you solve that equation, you can rearrange the gas law equation PV=nRT, where n is the number of moles of CO2 produced (moles of product!), and solve for V. [R=0.0821 (atm*L)/(mol*K)]

 

So, you end up with V=nRT/P, and if its properly set up, the units will cancel until you're left with just volume