How much 4 M KOH is required to react with 3.845g of Al metal?

Answer 1

First find the moles of Al. The relative atomic mass of Al equals Ar = 27, so

n = m/Ar = 3.845/27 = 0.142 mol (approx)

From the chem. equation:

2Al + 6KOH --> 2K3AlO3 + 3H2

2 mol of Al reacts with 6 mol KOH, so
0.142 mol Al reacts with 6*0.142/2 = 0.427 mol KOH

At last find the volume of the KOH solution:

n = C*V, V = n/C, V = 0.427/4, V = 0.107 L or 107 mL (approx)