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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?

Respuesta :

malika8888
Ca(OH)2+2HNO3=Ca(NO3)2+2H20

[tex]M = \frac{n}{V} \: [/tex]


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

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

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