Question

Gaseous butane CH3CH22CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 4.07 g of butane is mixed with 26. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 3 significant digits.

Answer 1

butane = C4H10

the balanced equation is as follows

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

moles = mass / molar mass

for C4H10 => 4.07g / 58.12g/mol => 0.07 moles

for O2 => 26 g / 32 g/mol => 0.8125 moles

2mol C4H10 -----------------> 8 mol CO2

0.07 mol -------------------> ?

=> 0.07 * 8 / 2 => 0.28 moles

13 moles O2 ----------------------> 8 mol CO2

0.8125 mol -------------------------> ?

=> 0.8125 * 8 / 13 => 0.5 moles

limiting reactant is C4H10

mass of CO2 formed => 0.28 * 44 => 12.325 grams