How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem.

2 Al (s) + 3 Fe(NO3)2 (aq) 3 Fe (s) + 2 Al(NO3)2 (aq)

Question

How many grams of iron metal do you expect to be produced when 305 grams of a 75.5 percent by mass iron(II) nitrate solution reacts with excess aluminum metal? Show all of the work needed to solve this problem. 

2 Al (s) + 3 Fe(NO3)2 (aq)   3 Fe (s) + 2 Al(NO3)2 (aq)

anonymous · Answer

First find your mass of iron nitrate
$$305* (75.5/100)= 230g$$
Then you would need the molar mass of the iron nitrate compound to calculate the moles of iron nitrate
230/180=1.28mol
Now, by looking at the stoichometric coefficients you can determine that for every 1mol of iron nitrate, you have 1 mol of iron. So you have 1.28 moles of iron. 
Now you simply multiply by the molar mass of iron 
$$1.28*55.8=71.4$$
The answer is 71.4g