Question

Can someone check my work?
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) yields 3 Fe (s) + 2 Al(NO3)2 (aq)

Answer 1

305 g Fe(NO3)2 * 75.5 g Fe(NO3)2/100 g Fe(NO3)2 * 1 mol Fe(NO3)2 = 1.26 mol Fe(NO3)2
1.26 mol Fe(NO3)2 * 3 mol Fe/2 mol Al = 1.89
1.89 * 55.845 g Fe/1 mol Fe = 105.55 g Fe

Answer 2

The molar ratio you used in step 2 is wrong.

no need to use aluminum

Answer 3

Would I use Iron (II) nitrate? Which would make the answer 70.36

Answer 4

i'm not sure what you're referring to

Answer 5

your error is in step 2

1.26 mol Fe(NO3)2 * 3 mol Fe/2 mol Al = 1.89

Answer 6

The mole ratio, what would I use to determine the moles of Iron metal?

Answer 7

yep that would be correct.

Answer 8

Thanks bud.

Answer 9

np