anyolsDr.Dariush Vosooghi 1. In the following chemical reaction: Al203 (s)+ 3C (

Question

anyolsDr.Dariush Vosooghi 1. In the following chemical reaction: Al203 (s)+ 3C (s)3CO(g) + 2Al (s) Starting with 3.50 gram of Al Os and 7.00 atoms of C find a. The limiting reagent b. How much (grams) Al would be produced? C. How much (grams) the excess reactant will be left over? ATOMIC Weight of Al 27.00 g/mole ATOMIC Weight of C = 12.00 g/mole ATOMIC Weight of O = 16.00 g/mole 2. In the following reaction; CuSO4·5H20 (s)+ Zn (s) ZnS049 + 5H20 (I) + Cu(s) 1.25 grams of CuS04.5H20 and 0.50 grams of Zn combine producing 0.25 grams of Cu. Find the percent yield of formation of Copper. Use the following atomic weight: Cu = 64.0 g/mol Zn = 65.0 g/mol s-32.0 g/mol O = 16.0 g/mol H = 1.0 g/mol

Explanation / Answer

1.

Al2O3 (s) + 3 C (s) ---------> 3 CO (g) + 2 Al (s)

Moles of Al2O3 = mas / molar mass = 3.50 / 102 = 0.0343 mol

Moles of C = 7.00 / (6.022 * 1023 ) = 1.16 * 10-23 mol

(a)

C is limiting reagent.

(b)

3 mol of C produces 2 mol of Al

Then,

1.16 * 10-23 mol of C produces 2 * 1.16 * 10-23 / 3 = 7.75 * 10-24 mol of Al

Mass of Al = moles * molar mass = 7.75 * 10-24 * 27.0 = 2.09 * 10-22 g. of Al

(c)

3 mol of C required 1 mol of Al2O3

then,

1.16 * 10-23 mol of C requries 3.48 * 10-23 mol of Al2O3

SO, left over excess reagent = 0.0343 - (3.48 * 10-23) = 0.0343 (approxi)

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