what is the mass of alum that can be prepared from 1.340g of Al.
2Al+2KOH+4H2SO4+22H2O ---> 3H2 + 2KAl(SO4)2 * 12H2O
The Attempt at a Solution
given Al is the limiting reagent. Find the moles of Al using n=m/M
I find Al moles to be 0.049662mol.
Now multiply this by the molar ratio of 2/2 (which is 1) to find the moles of alum produced.
now calculate for the mass, m = nM. The molar mass of alum is 474.3g/mol. I find the mass of alum to be 23.55g.
Am I right?