Question

A 50.0 mL sample if 0.0500 M Ca(OH)2 is added to 10.0 mL of 0.0200M HNO3. How many miles of excess acid or base are present?

Answer 1

Ca(OH)2+2HNO3=Ca(NO3)2+2H20


`M = (n)/(V) \:`



`0.05= (x)/(0.05) \\ x =0.0025 \: mol \: of \: Ca(OH)2 \:`

`0.02 = (y)/(0.01) \\ y = 0.002 \: mol \: of \: HNO3 \:`

HNO3 acid is limit
so,
0.0005 mol of Ca(OH)2 is excess
or
0.037 g of Ca(OH)2 is excess