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atroni [7]
3 years ago
7

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 precipita

te. (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?
Chemistry
1 answer:
mestny [16]3 years ago
4 0

Answer:

The answer is:

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

(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 AgNO_3,

= 30.0 ml

or,

= 0.030 L

Molarity of AgNO_3,

= 0.050 M

Moles of NaCl will be:

= \frac{Given \ mass}{Molar \ mass}

= \frac{0.0860}{58.44}

= 0.00147 \ mol

now,

Moles of AgNO_3 will be:

= Molarity\times Volume

= 0.050\times 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\times \frac{1}{1}\times \frac{143.32}{1}

= 0.211 \ g \ AgCl

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6 0
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For the reaction below, Kp = 1.16 at 800.°C. CaCO3(s) equilibrium reaction arrow CaO(s) + CO2(g) If a 25.0-g sample of CaCO3 is
goblinko [34]

Answer:

76.0%

Explanation:

Let's consider the following reaction.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

At equilibrium, the equilibrium constant Kp is:

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0.190molCO_{2}.\frac{1molCaCO_{3}}{1molCO_{2}} .\frac{100.09gCaCO_{3}}{1molCaCO_{3}} =19.0gCaCO_{3}

The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:

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The given units are the units of masses which are interconvertable.

It is given that:

1 lb = 16 oz

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