If 124 grams of Aluminum reacts with 601 grams of Fe2O3 as per the following reaction Al + Fe2O3 ------------------> Al2O3 + Fe Calculate the mass of Al2O3 formed Indicate which is the limiting reagent.
Added by Carmen R.
Step 1
85 g/mol O: 16.00 g/mol (3 atoms) Fe2O3 = 2(55.85) + 3(16.00) = 159.70 g/mol Show more…
Show all steps
Your feedback will help us improve your experience
Taimoor Shabbir and 96 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Consider the reaction: 4 Al2O3 + 9 Fe -> 3 Fe3O4 + 8 Al Calculate the amount (in moles) of aluminum, Al, produced when 5.00 moles of Al2O3 and 11.00 moles of Fe are mixed. Identify the limiting reagent.
Anna D.
The production of Aluminum oxide (Al2O3) is usually found present in cosmetics. It is possible with the reaction between aluminum and Iron (III) oxide (Fe2O3). To determine the limiting reactant in this reaction, calculate the moles of Al2O3 produced for each reactant if 250 g of Al is reacted with 801 g of Fe2O3. (Molar mass: Al = 26.98 g/mol, Fe2O3 = 159.69 g/mol)
David C.
Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O2(g)-2 Al2O3(s) a. 1 mol Al, 1 mol O2 b. 4 mol Al, 2.6 mol O2 c. 16 mol Al, 13 mol O2 d. 7.4 mol Al, 6.5 mol O2
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD