Oh sorry I forgot the arrows!
Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is given below.
Fe2O3(s) + 2 Al(s) -->2 Fe(l) + Al2O3(s)
What masses of iron(III) oxide and aluminum must be used to produce 30.0 g iron? What is the maximum mass of aluminum oxide that could be produced?
1. Convert 30 g Fe to moles. moles = grams/molar mass.
2. Using the coefficients in the balanced equation, convert moles Fe to moles Fe2O3.
3. Now convert moles Fe2O3 to grams. g = moles x molar mass.
4. Same process to determine g Al needed.
Post your work if you get stuck.
Ohh okay THANK YOU so much drbob222!
NFe2O3/1=0.26/2?
To determine the masses of iron(III) oxide and aluminum required to produce 30.0 g of iron, you need to use stoichiometry and the balanced chemical equation. Here's how you can do it:
1. Start by writing down the balanced chemical equation:
Fe2O3(s) + 2 Al(s) --> 2 Fe(l) + Al2O3(s)
2. Determine the molar masses of the compounds:
Iron(III) oxide (Fe2O3): 2(55.845 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
Aluminum (Al): 26.98 g/mol
Iron (Fe): 55.845 g/mol
Aluminum oxide (Al2O3): 2(26.98 g/mol) + 3(16.00 g/mol) = 101.96 g/mol
3. Use the molar masses to convert grams to moles:
Moles of iron(III) oxide (Fe2O3) = 30.0 g / 159.69 g/mol
Moles of iron(III) oxide (Fe2O3) = 0.1878 moles
4. Apply the stoichiometry of the balanced equation to determine the moles of aluminum required:
Moles of aluminum (Al) = (2 mol Al / 1 mol Fe2O3) * 0.1878 mol Fe2O3
Moles of aluminum (Al) = 0.3756 moles
5. Convert moles of aluminum to grams:
Mass of aluminum (Al) = 0.3756 moles * 26.98 g/mol
Mass of aluminum (Al) = 10.12 g
So, to produce 30.0 g of iron, you would need 0.1878 moles of iron(III) oxide and 10.12 g of aluminum.
To find the maximum mass of aluminum oxide that could be produced, you need to calculate the mass of aluminum oxide produced from the given mass of aluminum used:
1. Use the stoichiometry of the balanced equation to determine the moles of aluminum oxide produced:
Moles of aluminum oxide (Al2O3) = (1 mol Al2O3 / 2 mol Al) * 0.3756 moles Al
Moles of aluminum oxide (Al2O3) = 0.1878 moles
2. Convert moles of aluminum oxide to grams:
Mass of aluminum oxide (Al2O3) = 0.1878 moles * 101.96 g/mol
Mass of aluminum oxide (Al2O3) = 19.16 g
Therefore, the maximum mass of aluminum oxide that could be produced is 19.16 g.