molar mass of Al is 27 g/mol
we used .506 g/27 g/mol
= .0187 mols of Al
for every mol of Al we get a mol of K[Al(OH)4]
so .0187
A student reacts 0.506 grams of aluminum with excess KOH and H2O according to reaction one from the lab manual.
2Al + 2KOH +6H2O ---> 2K[Al(OH)4]+3H2
Enter your answers in numerical format.
How many moles of aluminum was reacted?
How many moles of K[Al(OH)4] will theoretically be produced?
1 answer
1 answer