Question

 

liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water.

suppose 38. G of hexane is mixed with 170. G of oxygen calculate the minimum mass of hexane that could be left over by the chemical reaction

round your answer to 2 significant digits

Answer 1

Balance the chemical equation first,

2CH₃(CH₂)₄CH₃ + 19O₂ -----------> 12 CO₂ + 14H₂O

moles of hexane = 38 gm / 86.18 g/mol = 0.4409 Moles

moles of O₂ = 170 gm / 32 gm/mol = 5.3125 Moles

2 Moles of CH₃(CH₂)₄CH₃ requires 19 moles of O₂

Therefore, 1 mole of CH₃(CH₂)₄CH₃ will require 19/2 moles of O₂

Hence, 0.4409 Moles of CH₃(CH₂)₄CH₃ will require 0.4409 * 19/2 = 4.188 moles of O₂

and we are given with 5.3125 Moles of O₂ that means CH₃(CH₂)₄CH₃ is the limiting reagent.

Now, hexane is the limiting reagent that means it will completely used in the reaction and at the end of the reaction no Mass of Hexane will be left.