Can someone help me answer this please:
Mass of aluminum metal used ---> 0.469 g
Mass of alum obtained --> 4.51 g
Molar mass of Aluminum used ---> 0.017 moles Al
Molar mass of alum (potassium aluminum sulfate) , KAl(SO4)2 * 12H2O ---> 474.41 g/mole
Noting that one aluminum atom is in each formula unit of alum, we know that each mole of aluminum will give rise to one mole of alum. Based on the number of moles of aluminum you used, calculate the maximum number of grams of alum that you produce, assuming that aluminum is the limiting reagent: ______ g alum.
This is what I got --> 8.06 g alum
0.017 moles Al x (1 mole Alum / 1 mole Al) x (474.4 g Alum/ 1 mole Alum) = 8.06 g alum
Not too sure about this answer...