How many grams of iron metal do you expect to be produced when 245 grams of an 80.5 percent by mass iron (ii) nitrate solution react with excess aluminum metal? show all of the work needed to solve this problem. 2al (s) + 3fe(no3)2 (aq) yields 3fe (s) + 2al(no3)3 (aq)

2Al + 3Fe(NO₃)₂ = 3Fe + 2Al(NO₃)₃

m=245 g
w=0.805 (80.5%)
M{Fe(NO₃)₂}=179.857 g/mol
M(Fe)=55.847 g/mol

1. the mass of salt in solution is:
m{Fe(NO₃)₂}=mw

2. the proportion follows from the equation of reaction:
m(Fe)/3M(Fe)=m{Fe(NO₃)₂}/3M{Fe(NO₃)₂}

m(Fe)=M(Fe)m{Fe(NO₃)₂}/M{Fe(NO₃)₂}

m(Fe)=M(Fe)mw/M{Fe(NO₃)₂}

m(Fe)=55.847*245*0.805/179.857= 61.24 g