Question

A 25.0-mL sample of a 1.20 M potassium chloride solution is mixed with 15.0 mL of a 0.900 M lead(II) nitrate solution and this precipitation reaction occurs: 2 KCl(aq)+Pb(NO3)2(aq)→PbCl2(s)+2 KNO3(aq) The solid PbCl2 is collected, dried, and found to have a mass of 2.45 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Answer 1

Answer:

Limiting reagent = lead(II) nitrate

Theoretical yield = 3.75435 g

% yield = 65.26 %

Explanation:

Considering:

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

Or,

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

Given :

For potassium chloride :

Molarity = 1.20 M

Volume = 25.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of potassium chloride :

$Moles=1.20 \times {25.0\times 10^{-3}}\ moles$

Moles of potassium chloride  = 0.03 moles

For lead(II) nitrate :

Molarity = 0.900 M

Volume = 15.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 25.0×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=0.900 \times {15.0\times 10^{-3}}\ moles$

Moles of lead(II) nitrate  = 0.0135 moles

According to the given reaction:

$2KCl_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbCl_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium chloride react with 1 mole of lead(II) nitrate

1 mole of potassium chloride react with 1/2 mole of lead(II) nitrate

0.03 moles potassium chloride react with 0.03/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.015 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) nitrate is limiting reagent. (0.0135 < 0.015)

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

1 mole of lead(II) nitrate gives 1 mole of lead(II) chloride

0.0135 mole of lead(II) nitrate gives 0.0135 mole of lead(II) chloride

Molar mass of lead(II) chloride = 278.1 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.0135 × 278.1 g = 3.75435 g

Theoretical yield = 3.75435 g

Given experimental yield = 2.45 g

% yield = (Experimental yield / Theoretical yield) × 100 = (2.45/3.75435) × 100 = 65.26 %

Answer 2

Limiting reactant for the given reaction is lead (II) nitrate, theoretical yield is 3.75435 g and percent yield is 65.26 %.

What is limiting reactant?

Those reactant which is present in less quantity as compare to the another reactant is known as limiting reactant.

Balanced given chemical reaction is:

2KCl₍aq₎ + Pb(NO₃)₂₍aq₎ → PbCl₂₍s₎ + 2KNO₃₍aq₎

Also given that molarity of KCl = 1.20 M

Volume of KCl = 25 mL or 0.025 L

Molarity of Pb(NO₃)₂ = 0.900 M

Volume of Pb(NO₃)₂ = 15mL or 0.015 L

(1). First we determine the limiting reactant:

From the reaction it is clear that 1 mole of Pb(NO₃)₂ is required to completely react with 2 mole of KCl. So, half mole of Pb(NO₃)₂ is react with 1 mole of KCl.

So, limiting reactant is lead (II) nitrate.

(2). Now we calculate the theoretical yield:

Formation of the product in any chemical reaction is governed by the limiting reactant. Hence it is clear that 1 mole of lead (II) nitrate will produce 1 mole of PbCl₂.

We can calculate the moles from the equation of molarity i.e. M = n/V

or n = M × V

No. of moles of Pb(NO₃)₂ = 0.90 × 0.015 = 0.0135 mole

So, no. of moles of PbCl₂ = 0.0135 mole

We can calculate produced mass of PbCl₂ by using the formula n = W/M

Where, W = weight

M = molar mass

W = n × M

W = 0.0135 × 278.1 = 3.75435 g.

(3). Percent yield can be calculated as:

percent yield = (experimental yield / theoretical yield) × 100

% yield = (2.45 / 3.75435) × 100 = 65.26%

Therefore, lead (II) nitrate is the limiting reactant, 3.75435 g is the theoretical yield of the precipitate and 65.26% is the percent yield.

To know more about limiting reactant, visit the below link:

https://brainly.com/question/14222359