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Fe₂O3 +2A1→→→→2Fe + Al₂O3

How many grams of Al are needed to completely react with 135 g
Fe₂O3?

How many grams of Al₂O3 can form when 23.6 g Al react with excess
Fe₂O3?

How many grams of Fe₂O3 react with excess Al to make 475 g Fe?

How many grams of Fe will form when 97.6 g Al2O3 form?

1 Answer

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Answer:

To solve stoichiometry problems like these, we need to use balanced chemical equations and convert the given quantities into moles using their respective molar masses. Then, we can use the mole ratios from the balanced equation to find the moles of the unknown substance, and convert back to grams if necessary.

Fe₂O3 + 2Al → 2Fe + Al₂O3

How many grams of Al are needed to completely react with 135 g Fe₂O3?

First, we need to find the moles of Fe₂O3 in 135 g:

molar mass of Fe₂O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol

moles of Fe₂O3 = 135 g / 159.70 g/mol = 0.8459 mol

According to the balanced equation, 2 moles of Al react with 1 mole of Fe₂O3:

moles of Al = 2 × moles of Fe₂O3 = 2 × 0.8459 mol = 1.6918 mol

Finally, we can convert the moles of Al to grams using its molar mass:

molar mass of Al = 26.98 g/mol

mass of Al = moles of Al × molar mass of Al = 1.6918 mol × 26.98 g/mol ≈ 45.66 g

Therefore, 45.66 grams of Al are needed to completely react with 135 g Fe₂O3.

How many grams of Al₂O3 can form when 23.6 g Al react with excess Fe₂O3?

First, we need to find the moles of Al in 23.6 g:

moles of Al = 23.6 g / 26.98 g/mol ≈ 0.874 mol

According to the balanced equation, 2 moles of Al react with 1 mole of Al₂O3:

moles of Al₂O3 = 0.5 × moles of Al = 0.5 × 0.874 mol = 0.437 mol

Finally, we can convert the moles of Al₂O3 to grams using its molar mass:

molar mass of Al₂O3 = 101.96 g/mol

mass of Al₂O3 = moles of Al₂O3 × molar mass of Al₂O3 = 0.437 mol × 101.96 g/mol ≈ 44.64 g

Therefore, 44.64 grams of Al₂O3 can form when 23.6 g Al react with excess Fe₂O3.

How many grams of Fe₂O3 react with excess Al to make 475 g Fe?

First, we need to convert the given mass of Fe to moles:

molar mass of Fe = 55.85 g/mol

moles of Fe = 475 g / 55.85 g/mol ≈ 8.504 mol

According to the balanced equation, 2 moles of Al react with 1 mole of Fe:

moles of Al = 0.5 × moles of Fe = 0.5 × 8.504 mol = 4.252 mol

Since Al is in excess, it will not be completely consumed by the reaction. However, we can use the moles of Al to find the moles of Fe₂O3 needed:

moles of Fe₂O3 = 0.5 × moles of Al = 0.5 × 4.252 mol = 2.126 mol

Finally, we can convert

Step-by-step explanation:

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