posted by Anonymous on .
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) (4 points)
You need to learn how to write arrows. --> or ==> or >>> will do it. Technically, the equation you wrote was incorrect since it doesn't separate the reactants from the products.
305 g sample Fe(NO3)2 has how much Fe(NO3)2? That's 305 x 0.755 = ?
mols Fe(NO3)2 = grams/molar mass
Use the coefficients in the balanced equation to convert mols Fe(NO3)2 to mols Fe.
Now convert mols Fe to grams Fe. g = mols x atomic mass Fe.