A sample of 0.0860 g of sodium chloride is added to 30.0 mL of 0.050 M silver nitrate, resulting in the formation of a precipitate. (a) Write the molecular equation for the reaction. (b) What is the limiting reactant in the reaction? (c) How many grams of precipitate potentially form?

The answer is:

(a)

(b) NaCl

(c) 0.211 g

Explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of ,

= 30.0 ml

or,

= 0.030 L

Molarity of ,

= 0.050 M

Moles of NaCl will be:

=

=

=

now,

Moles of will be:

(a)

The reaction is:

⇒

(b)

1 mole of NaCl react with,

= 1 mol of

0.0015 mol needs,

=

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

=

=

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