Question

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?

Answer 1

The answer is:

(a)
`NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)`

(b) NaCl

(c) 0.211 g

Step-by-step explanation:

Given:

The mass of NaCl,

= 0.0860 g

The molar mass of NaCl,

= 58.44 g/mol

The volume of
`AgNO_3`,

= 30.0 ml

or,

= 0.030 L

Molarity of
`AgNO_3`,

= 0.050 M

Moles of NaCl will be:

=
`(Given \ mass)/(Molar \ mass)`

=
`(0.0860)/(58.44)`

=
`0.00147 \ mol`

now,

Moles of
`AgNO_3` will be:

`= Molarity* Volume`

`= 0.050* 0.030`

`=0.0015 \ mol`

(a)

The reaction is:

⇒
`NaCl(aq)+AgNO_3(aq) \rightarrow AgCl(r) +NaNO_3 (aq)`

(b)

1 mole of NaCl react with,

= 1 mol of
`AgNO_3`

0.0015 mol
`AgNO_3` needs,

=
`0.00150 \ mol \ NaCl`

Available mol of NaCl < needed amount of NaCl

So,

The limiting reagent is "NaCl".

(c)

The precipitate formed,

=
`0.00147* (1)/(1)* (143.32)/(1)`

=
`0.211 \ g \ AgCl`