A mixture of hydrocarbons contains 38.3% hexane, C6H14, 13.9% octane, C8H18, and 47.8% decane, C10H22. The mixture is combusted in an excess of oxygen. What mass of CO2 is generated from the combustion of 19.3 kg of the mixture

52.206 kg

Explanation:

From the given information:

Mass of hexane C6H14 =

= 7391.9 g

Mass of octane C8H18 =

= 2682.7 g

Mass of decane C10H22 =

= 9225.4 g

However, recall that:

number of moles of an atom = mass/molar mass

∴

For hexane, no of moles = 7391.9 g/86.18 g/mol

= 85.77 moles

For octane, no of moles = 2682.7 g/114.23 g/mol

= 23.49 moles

For decane, no of moles = 9225.4 g/142.29 g/mol

= 64.84 moles

Therefore:

number of moles of CO2 produced = (6 × 85.77)+(23.49)+(10×64.84) moles

= 1186.51 moles

Finally, the mass of CO2 produced is:

= 1186.51 mol × 44 g/mol

= 52206.44  g

= 52.206 kg

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm