Round-Robin: In the Round-Robin: In the following problems, each member of the g

Question

Round-Robin: In the Round-Robin: In the following problems, each member of the group will be responsible for a step in the problem solving process. The first person will read the problem aloud, do Step 1, and explain the answer to the group. If the group agrees, the second person will do step two, and so-on until the problem is completed. Each member should initial all papers beside the step 5/he attempts Powdered aluminum metal reacts with iron(III) oxide in a process 2 Al(s) + Fe_20_3(s) implies 2 Fe(I) + AI_2O_3(S) Consider a reaction in which 113.32 grams of aluminum is used up. How many grams of molten iron and of aluminum oxide can be made?. STEP #1: Describe, in words, what the balanced chemical equation means. For example, does that "for every 2 grams of aluminum used up, 1 gram of iron(III) oxide is formed? STEP #2: Into what units do you first need to change the grams of aluminum? Do the calculation, STEP #3: Describe how the "2" in front of Al(s) is different from the 113.32 g Al(s) from the problem. STEP #4: Calculate the number of aluminum atoms in 113.32 g Al.

Explanation / Answer

step 4 : moles of Al = mass / molar mass = 113.32 / 27 = 4.197

number of Al atoms = 4.197 x (6.023 x 10^23) = 2.53 x 10^24 atoms

step 5:

2 moles Al -------------------------> 1 mole Al2O3

4.197 moles Al --------------> 4.197 /2 mole Al2O3 = 2.0985 mole Al2O3

Al2O3 molecules = 2.0985 x (6.023 x 10^23) = 1.26 x 10^24 molecules

step 6 :

Al2O3 moles = 2.0985

Al2O3 mass = 2.0985 x 101.96 = 213.96 g

step 7 :

2 moles Fe = 1 mole Al2O3

moles of Fe = 1/2 x 2.0985 = 1.049 moles of Fe

moles of Fe = 1.049

mass of Fe = 1.049 x 55.845 = 58.60 g

  

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