Question

Liquid hexane (CH₃(CH₂)₄CH₃) will react with gaseous oxygen (O₂ ) to produce gaseous carbon dioxide (CO₂ ) and gaseous water (H₂ O). Suppose 6.03 g of hexane is mixed with 35. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Molar mass of C6H14,

MM = 6*MM(C) + 14*MM(H)

= 6*12.01 + 14*1.008

= 86.172 g/mol

mass(C6H14)= 6.03 g

use:

number of mol of C6H14,

n = mass of C6H14/molar mass of C6H14

=(6.03 g)/(86.17 g/mol)

= 6.998*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 35.0 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(35 g)/(32 g/mol)

= 1.094 mol

Balanced chemical equation is:

2 C6H14 + 19 O2 ---> 12 CO2 + 14 H2O

2 mol of C6H14 reacts with 19 mol of O2

for 6.998*10^-2 mol of C6H14, 0.6648 mol of O2 is required

But we have 1.094 mol of O2

so, C6H14 is limiting reagent

Since C6H14 is limiting reagent, all of C6H14 will react

mass of C6H14 left over = 0.00 g

Answer: 0.00 g