Using the equation, #Fe_2O_3 + 2Al - 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?

Question

Using the equation, #Fe_2O_3 + 2Al -> 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?

Natalie Anton · Accepted Answer

#135g Fe_2 O_3 * (1 mol Fe_2 O_3) / (159.69g Fe_2 O_3) * (2 mol Al) / (1 mol Fe_2 O_3) * (26.98g Al) / (1 mol Al) = 45.6g Al#

You must first multiply the mass of iron (III) oxide by the inverse of its molar mass in order to convert it to moles of iron (III) oxide because the coefficients in the equation provide a molar ratio between the chemical species in the equation.

Then, by looking at the relationship between the moles of iron (III) oxide and aluminum, you can determine the number of moles of aluminum per mole of #Fe_2 O_3#.

Finally, convert the number of moles of aluminum to grams of aluminum by multiplying by its molar mass. After unit cancellation, you can determine that 45.6 g Al is necessary to completely react with 135 g #Fe_2 O_3#.

Axel Alvarado · Answer

undefined First, calculate the molar mass of Fe₂O₃ and Al₂O₃.

Fe₂O₃:
Fe: 55.85 g/mol * 2 = 111.7 g/mol
O: 16.00 g/mol * 3 = 48.00 g/mol
Total: 159.7 g/mol

Al₂O₃:
Al: 26.98 g/mol * 2 = 53.96 g/mol
O: 16.00 g/mol * 3 = 48.00 g/mol
Total: 101.96 g/mol

Now, calculate the moles of Fe₂O₃:
135 g Fe₂O₃ / 159.7 g/mol = 0.846 moles Fe₂O₃

According to the balanced equation, the molar ratio of Fe₂O₃ to Al is 1:2.
So, moles of Al needed = 0.846 moles Fe₂O₃ * (2 moles Al / 1 mole Fe₂O₃) = 1.692 moles Al

Now, calculate the grams of Al needed:
1.692 moles Al * 26.98 g/mol = 45.66 grams of Al

Using the equation, #Fe_2O_3 + 2Al -&gt; 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?

Using the equation, #Fe_2O_3 + 2Al -> 2Fe + Al_2O_3#, how many grams of #Al# are needed to completely react with 135 grams of #Fe_2O_3#?