Based on the diagram above, nitromethane has a chemical formula of CH _ _

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Nitromethane, sometimes shortened to just Nitro, is an organic compound with the chemical formula CH

3NO

2. It is the simplest organic nitro compound. It is a polar liquid commonly used as a solvent in a variety of industrial applications such as in extractions, as a reaction medium, and as a cleaning solvent. As an intermediate in organic synthesis, it is used widely in the manufacture of pharmaceuticals, pesticides, explosives, fibers, and coatings.[12] Nitromethane is used as a fuel additive in various motorsports and hobbies, e.g. Top Fuel drag racing and miniature internal combustion engines in radio control, control line and free flight model aircraft.

Explanation:

Preparation

Nitromethane is produced industrially by combining propane and nitric acid in the gas phase at 350–450 °C (662–842 °F). This exothermic reaction produces the four industrially significant nitroalkanes: nitromethane, nitroethane, 1-nitropropane, and 2-nitropropane. The reaction involves free radicals, including the alkoxyl radicals of the type CH3CH2CH2O, which arise via homolysis of the corresponding nitrite ester. These alkoxy radicals are susceptible to C—C fragmentation reactions, which explains the formation of a mixture of products.[12]

Laboratory methods

It can be prepared in other methods that are of instructional value. The reaction of sodium chloroacetate with sodium nitrite in aqueous solution produces this compound:[13]

ClCH2COONa + NaNO2 + H2O → CH3NO2 + NaCl + NaHCO

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