The reaction is: Fe2O3 (s) + 2Al (s) → 2Fe (l) + Al2O3 (s)
a.How many moles of aluminum must react to produce 2.4 mol of Al2O3?
b. What mass, in grams, of iron (III) oxide reacted to produce 4.5 mol of Al2O3?
c. If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced?
d. If 3.6 g of molten iron is produced during the reaction, how many grams of Al2O3 is produced

Confused, pretty sure I did this wrong so need help

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-11-15T15:26:14+01:00

There are 4.8 moles of Al2O3 produced when 2.4 mol of Fe2O3 reacts with iron.

a)The equation of the reaction is; Fe2O3 (s) + 2Al (s) → 2Fe (l) + Al2O3 (s)

From the reaction equation;

2 moles of Al produces 1 moles of Al2O3

x moles of Al produces 2.4 moles of Al2O3

x = 2 moles × 2.4 moles/1 moles

x = 4.8 moles

b) 1 mole of Fe2O3 produces 1 moles of Al2O3

4.5 moles of Fe2O3 produces 4.5 moles of Al2O3

c) Number of moles in 3.6 g of Al = 3.6 g/27 g/mol = 0.133 moles

If 2 moles of Al yields 1 mole of Al2O3

0.133 moles Al yields 0.133 moles × 1 mole/ 2 moles

= 0.0665 moles

Mass of Al2O3 = 0.0665 moles × 102g/mol

= 6.8 g

d) Number of moles of iron = 3.6 g/56 g/mol = 0.064 moles

From the reaction equation;

1 mole of Al2O3 is produced when 2 moles of iron is produced

x moles of Al2O3 is produced when 0.064 moles is produced

x = 1 mole × 0.064 moles/ 2 moles

x = 0.032 moles

Mass of Al2O3 produced = 0.032 moles × 102g/mol

= 3.3 g

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