Question

Liquid hexane CH3CH24CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 1.7 g of hexane is mixed with 4.44 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

molar mass of hexane = 85 g/mol

mass of hexane = 1.7 g

number of moles of hexane =  

=

= 0.02 moles

molar mass of oxygen = 16 g/mol

mass of oxygen = 4.44 g

number of moles of oxygen =  

=

= 0.2775 moles

alkanes reacting with oxygen to give carbon dioxide and water is called a combustion reaction

Alkanes are of the form C_nH_(2n+2)

The equation for complete combustion reaction is

C_nH_2n+2 + (3n/2 + 1/2) O₂ (n + 1) H₂O + n CO₂

since the alkane is hexane n=6

therefor one mole of hexane needs (36/2 + 1/2) = 9.5 moles of oxygen for complete combustion

number of moles needed for 0.2 moles of hexane = 0.2 9.5

= 1.9 moles

since the number of moles of oxygen are less than 1.9 moles hexane wont undergo compete combustion

we now will find out number of moles of hexane that 0.2775 moles of oxygen can burn

since one mole of hexane needs 9.5 moles of oxygen for complete combustion one mole of oxygen reacts with moles of hexane

number of moles of hexane that can react with 0.2775 moles of oxygen = moles

= 0.029 moles

1 mole of alkane produces n moles of carbon dioxide

Therefore 1 mole of hexane produces 6 moles of carbon dioxide

number of moles of carbon dioxide produced by 0.029 moles of oxygen = 6 0.029

= 0.174 moles

mass of carbon dioxide produced = number of moles of carbon dioxide molar mass of carbon dioxide

= 0.174 moles 44 g/mol

= 7.656 g

= 7.66 g ( rounding off to three significant digits )