Question

Using this reaction: 4Al(s) + 3O2(g) ----> 2Al2O3(s)

a. Determine the limiting reactant and the maximum yield if 4.70g Al reacts with 5.00L of O2 at STP. ( I got Al as the limiting reactant and the maximum yield of .087mol.)

b. If the actual yield is 8.03 grams, what is the % yield. ( I got 90.4%) Did I do these right?

Answer 1

First. Verify that chemical reaction is balanced

a) MW (Al)= 26,98g/mol

MW (O₂)=32g/mol

MW(Al₂O₃)=101,96g/mol

-----Al moles ---------------

------O₂ moles--------------

STP=( T=273,15K; P=1atm ; R=0,082 atm.L/K.mol)

--------------------Determining reactant limiting

Is necessary evaluate stoichiometric of the reaction: 4Al and 3 O₂

   _and

_{Al is limiting reactant}

_Yield:

  of Al₂O₃

You answer is correct!

b) maximum yield is 0,087 mol Al₂O₃

for 8,03g of product

of Al₂O3

  %

Check your answer.

It was a pleasure!