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Lapatulllka [165]
3 years ago
13

Write balance complete molecular equation, ionic equation, and net ionic equations for the reactions that occur when each of the

following solutions are mixed. Identify the spectator ion or ions in each reaction: a) Pb(NO3)2 and Na2S
Chemistry
1 answer:
AveGali [126]3 years ago
3 0

<u>Answer:</u> The complete molecular, ionic, and net ionic equations are given below. The spectator ions are sodium and nitrate ions.

<u>Explanation:</u>

The ionic equation is defined as the equation in which all the substances that are strong electrolytes present in an aqueous state and are represented in the form of ions.

The net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

The balanced molecular equation for the reaction of lead (II) nitrate and sodium sulfide follows:

Pb(NO_3)_2(aq)+Na_2S(aq)\rightarrow PbS(s)+2NaNO_3(s)

The ionic equation follows:

Pb^{2+}(aq)+2NO_3^-(aq)+2Na^+(aq)+S^{2-}(aq)\rightarrow PbS(s)+2Na^+(aq)+2NO_3^-(aq)

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Pb^{2+}(aq)+S^{2-}(aq)\rightarrow PbS(s)

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A 57.0 mL sample of a 0.120 M potassium sulfate solution is mixed with 35.5 mL of a 0.118 M lead(II) acetate solution and the fo
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Answer:

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

Moles=0.120 M \times {57.0\times 10^{-3}}\ moles

Moles of potassium sulfate  = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

Moles=0.118 \times {35.5\times 10^{-3}}\ moles

Moles of lead(II) acetate  = 0.004189 moles

According to the given reaction:

K_2SO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow 2KC_2H_3O_2_{(s)}+PbSO_4_{(aq)}

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

The formation of the product is governed by the limiting reagent. So,

1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %</u>

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