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29 July, 13:45

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.

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  1. 29 July, 16:11
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    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
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