Question

9. Certain race cars use methanol (CH3OH; also called wood alcohol) as a fuel. Methanol has a molecular mass of 32.0 g/mol and a density of 0.79 g/mL. The combustion of methanol occurs according to the following equation:

2CH3OH + 3O2 ? 2CO2 + 4H2O

In a particular reaction 2.00 L of methanol are reacted with 80.0 kg of oxygen.

a. What is the limiting reactant?

b. What reactant and how many grams of it are left over?

c. How many grams of carbon dioxide is produced?

NOTE: Please answer the (a,b,c) under its each draft.  

Answer 1

mass of methanol   = volume * density

                               = 2000*0.79   = 1580g

no of moles of methanol     = W/G.M.Wt

                                           = 1580/32   =49.375moles

no of moles of O2    = W/G.M.Wt

                               = 80000/32   = 2500moles

a.

2CH3OH + 3O2 ? 2CO2 + 4H2O

3 moles of O2 react with 2 moles of CH3OH

2500moles of O2 react with = 2*2500/3   = 1666.66 moles of CH3OH

CH3OH is limiting reactant

c

2 moles of CH3OH react with excess of O2 to gives 2 moles of CO2

49.375 moles of CH3OH react with excess of O2 to gives = 2*49.375/2   = 49.375 moles of CO2

mass of CO2   = no ofmoles * gram molar mass

                       = 49.375*44   = 2172.5g of CO2

b.

2 moles of CH3OH react with 3 moles of O2

49.375 moles of CH3OH react with = 3*49.375/2   = 74.0625 moles of O2 is required

The no of moles of O2 left over after complete the reaction = 1666.66-74.0625 = 1592.6 moles

The amount of O2 left over after complete the reaction = no of moles * gram molar mass

                                                                                         = 1592.6*32 = 50963.2 g