Calculate the mass, in grams, of CH3CH2NH2 produced from 32.9 g of CH3CH2I reacting with an excess of NH3 assuming a 70.0% yield.
To tackle any question like this, you should first write out the reaction formula (I would do this on the whiteboard). You can see from this that the ratio between the reactant and product is 1:1. You can also see that you have been given the mass and molecular formula of your starting compound. By adding up the masses of the individual atoms in the compound, you can calculate the molecular mass of the compound. You can then use the formula: moles = mass/molecular mass to determine the number of moles of CH3CH2I used in the reaction.
Now, using the ratio determined by writing out the reaction formula, you know that in a reaction with a 100% yield, you would generate the same number of moles of CH3CH2NH2 as CH3CH2I. However, in this case, the reaction yield is only 70.0%. This means the number of moles of CH3CH2NH2 made is only 70.0% of this number (0.7 x the number of moles). Finally, by rearranging the formula we used earlier to give mass = moles x molecular mass, you can calculate the mass of the product by again adding up the individual masses, and multiplying this number by the number of moles.
