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 react with excess aluminum metal? Show all of the work needed to solve this problem.

Question

Esmeralda

Chemistry

1 September, 10:31

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 react with excess aluminum metal? Show all of the work needed to solve this problem.

+2

Answers (1)

Know the Answer?

Ramon · Answer 1

The reaction will produce 71.5 g Fe.

Step 1. Write the balanced chemical equation

3Fe (NO_3) _2 + 2Al → 3Fe + 2Al (NO3) _3

Step 2. Calculate the mass of Fe (NO3) _2

Mass = 305 g solution * [75.5g Fe (NO_3) _2/100 g solution]

= 230.3 g Fe (NO_3) _2

Step 3. Calculate the moles of Fe (NO_3) _2

Moles of Fe (NO_3) _2

= 230.3 g Fe (NO_3) _2 * [1 mol Fe (NO_3) _2/179.85 g Fe (NO_3) _2]

= 1.280 mol Fe (NO_3) _2

Step 4. Calculate the moles of Fe

Moles of Fe = 1.280 mol Fe (NO_3) _2 * (3 mol Fe/3 mol Fe (NO_3) _2)

= 1.280 mol Fe

Step 5. Calculate the mass of Fe

Mass of Fe = 1.280 mol Fe * (55.845 g Fe/1 mol Fe) = 71.5 g Fe