1. The thermite reaction is performed using 8.6g Fe2O3 and 1.8g powdered Al meta

Question

1. The thermite reaction is performed using 8.6g Fe2O3 and 1.8g powdered Al metal/

a. Which reactant is the excess?

b. How many grams of excess reactant are left over?

c. How many grams of aluminum oxide can be produced?

2. The reaction of 5.0g of CuSo4 with excess sodium hydroxide produced 2.6g of Cu(OH)2. What percent yield of Cu(OH)2 was obtained?

3. Approximately 46.4 grams of sodium acetate (82.0 gm/mol) will dissolve in 100 grams of water at 20 degrees Celcius. If 2.75 mol NaCH3COO is dissolved in 355 g of water, the solution will be:

A. dilute

B. concentrated

C. saturated

D. supersaturated

4. Answer the following as true or false:

A. NaCl(aq) has a greater surface tension from H20

B. CH4 is more polarizable than CBr4

C. C12H6 is more viscous than C5H12

D. Compared to methanol, Bromine is more soluble in hexane.

E. Water rises in a capillary tube due to the fact that cohesive forces between the water molecules are stronger than the adhesive forces between water and the glass.

Explanation / Answer

The thermite reaction is given by:

2Al + Fe2O3 ==== Al2O3 + 2Fe

so we have to calculate the moles for the substances given with moles = mass / molar mass

Fe2O3 is 86 grams, molar mass 159.69 g/gmol

moles of Fe2O3 = 8.6 / 159.69 = 0.053854 moles

Al is 1.8 grams, molar mass 27 g/gmol

moles of Al = 1.8 / 27 = 0.0666 moles

from the balanced reaction we see that 2 moles of Al need 1 mole of Fe2O3 so if you multiply by 2 the moles of Fe2O3 you will get a value of 0.1077 moles you require of Al

since you have less than this the Al is the limiting reactant, The Fe2O3 is in excess

so we must make calculations according to the moles of Al available

so 2 moles of Al reacts with 1 mole of Al so 0.0666 moles will react with 0.0666 / 2 = 0.0333 moles of Fe2O3

moles of Fe2O3 in excess = 0.0538 - 0.0333 = 0.0205 moles of Fe in excess

grams of Fe2O3 are calculated with mass = moles * molar mass

mass = 0.0205 * 159.69 = 3.273 grams of Fe2O3 unreacted

2 moles of Al will produce only 1 of Al2O3 so 0.0666 moles of Al will produce 0.0333 moles of Al2O3

mass = moles * molar mass, molar mass of Al2O3 is 102 g/gmol

mass = 0.0333 * 102 = 3.3966 grams, this is the maximum ammount of Al2O3 that can be produced

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