Question

For this problem, identify the limiting reagent and calculate the grams of Al2O3 obtained in the reaction of 100. grams of aluminum with 70.0 grams oxygen. If 125 grams of Al2O3 is actually produced, what is the % yield. The equations are not balanced. They were balanced in Exercise B. Use those coefficients to do these calculations.

Al + O2→ Al2O3

What is the limiting reagent? ___________________

How many grams Al2O3 will be produced? ___________________

If 120 grams of Al2O3 is produced, what is the % yield. ____________________

Answer 1

Balanced reaction is 4Al + 3O2→ 2Al2O3, the reaction suggests 4 moles of Al reacts with 3 moles of O2 to produce 2 moles of Al2O3.

Atomic weights : Al = 27 g/mole , O= 16, molar mass of Al2O3= 2*27+3*16= 102 g/mole

moles of Al= 100/27= 3.7, O2= 70/32= 2.1875

as per the reaction, molar ratio of Al :O2= 4:3= 4/3 :1= 1.33 :1

actual molar ratio of Al:O2= 3.7:2.1875 = 3.7/2.1875: 2.1875/2.1875= 1.7:1

so excess is Al and all the O2 reacts. hence Oxygen is limiting reactant.

3 moles of O2 produces 2 moles of Al2O3

2.1875 moles of O2 produces 2.1875*2/3= 1.46 moles of Al2O3, mass of Al2O3= moles* molar mass = 1.46*102=148.92 gm This is the theoretical yiled of Al2O3.

actual yield = 100* actual yield/theoretical yield= 100*120/148.92 =80.6%