Question

28g of aluminum reacts with 45g oxygen

1) complete balanced equation?
2) what is the limiting reactant?
3) how many grams of product could be formed?
4) how many grams of excess would you predict?
5) if you actually measured 50g of product, what is the percent yield?

Answer 1

answer below:

Step-by-step explanation:

1) 4 Al + 3 O2 --> 2Al2O3.

2) Al mol= 28/27 = 1.037mol

O mol= 45/16 = 2.812mol

Al : O

4 : 2

x : 2.812

cross multiply

4 x 2.812/2 = 5.624 mol of Al

Al is limiting reactant

3) 28/5.246 = 5.34product can be formed (not that much sure)

4) 45/ 0.5185 = 86.79

86.79 - 5.34 = 81.45g

5) 50 / 81.45 x 100 = 61.39% ( i guess)

i hope this hleps you