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

$M = \frac{n}{V} \:$


$0.05= \frac{x}{0.05} \\ x =0.0025 \: mol \: of \: Ca(OH)2 \:$
$0.02 = \frac{y}{0.01} \\ y = 0.002 \: mol \: of \: HNO3 \:$

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