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)

Answer 1

We are given with 305 grams of a 75.5 percent by mass iron(II) nitrate solution to produce metal iron upon reaction with aluminum. The mass of iron(II) nitrate is 305 grams * 0.755 or equal to 230.275 grams. In this case, 1 mole of iron(II) nitrate is equal to 1 mole of iron. Thus we just divide the amount of iron(II) nitrate with molar mass of iron(II) nitrate and multiply with the molar mass of iron. The answer is 71.51 grams.

Answer 2

The balanced chemical reaction is:

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

The ratios or the coefficients between the reactants and the products will be used for further calculations.

305 g Fe(NO3)2 solution (.755) = 230.275 g Fe(NO3)2

230.275 g Fe(NO3)2 ( 1mol / 179.87 g) ( 3 mol Fe / 3 mol Fe(NO3)2) ( 55.85 g / mol) = 71.500 g Fe