Powdered aluminum is used in the thermite reaction for welding metals. what role does the powdered aluminum play in this reaction?

Powdered aluminum serves the reducing agent for this reaction. Aluminum powder is known as a powerful reducing agent. Iron(iii) ions would have greater electron affinity as compared to a metallic aluminum thus it would easily gain electrons from aluminum making it less positive in this case formed into a neutral atom. The chemical reaction is 

Fe2O3 + 2Al = Al2O3 + 2Fe. 

Oxygen in the oxide compound does not participate in this redox reaction, as it is not changed chemically. Fe2O3 is an ionic compound which has of Fe[3+] ions and O[2-] ions. So, the net ionic reaction would be

Fe3+ + Al + 3O2- = Al3+ + Fe + 3O2- 

The ionization energy of aluminum is greater than that of iron, so the said reaction would generate large amount of heat which is needed in welding.

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answer: 88.5%

explanation:

1) write the balanced chemical equation:

2h₂ + o₂ → 2h₂o

2) write the mole ratios:

2 mol h₂ : 1 mol o₂ : 2 moles h₂o

3) calculate the number of moles of each reactant:

formula: number of moles = mass in grams / molar massnumber of moles of h₂: 11.0 g / 2.016 g/mol = 5.46 molesnumber of moles of o₂: 95.0 g / 31.998 g/mol = 2.97 moles

4) determine the limiting reactant:

theoretical ratio: 1 mol o₂ / 2 mol h₂ = 0.5 o₂ / h₂actual ratio: 2.97 mol o₂ / 5.46 mol h₂ ≈ 0.54 o₂ / h₂hence, there is more o₂ moles than the needed to react with the h₂ moles given, so o₂ is in excess and h₂ is the limiting reactant.

5) calculate the theoretical yield using the limiting reactant:

proportion: 2 moles h₂ / 2 moles h₂o = 5.46 moles h₂ / xsolve for x: x = 5.46 moles of h₂oconvert moles to mass of h₂o: 5.46 mol × 18.015 g / mol = 98.4 g

6) calculate the percent yield:

percent yiled = (actual yield / theoretical yield) × 100percent yield = (87.0 g / 98.4 g) × 100 = 88.4% ← answer

note: the 0.1% difference of 88.4% with 88.5% is not relevant since the difference is in the last significant figure and it can be attributed to differences in the precision in the molar masses used.