How many grams of iron metal do you expect to be produced when 325 grams of an 87.5 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)

Question

How many grams of iron metal do you expect to be produced when 325 grams of an 87.5 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)

anonymous · Answer

i honestly dont know how to do this! and i have an exam coming up! can someone please help me!?!?!?!

wolfe8 · Answer

Hey welcome to OpenStudy. Calm down, we can help you, First, can you find the number of moles of Fe(NO3)2?

jfraser · Answer

If you know the total mass of the solution, and you know the percentage of iron (II) nitrate in the solution, you can easily find the mass of just the iron (II) nitrate. Then you can use the molar ratio from the balanced equation to find the reaction's yield.

anonymous · Answer

53.47