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Lead(ii) nitrate and ammonium iodide react to form lead(ii) iodide and ammonium nitrate according to the reaction pb(no3)2(aq)+2nh4i(aq)⟶pbi2(s)+2nh4no3(aq) what volume of a 0.290 m nh4i solution is required to react with 253 ml of a 0.680 m pb(no3)2 solution?

Respuesta :

The balanced chemical reaction between [tex]Pb(NO_{3})_{2}[/tex] and [tex]NH_{4}I[/tex] is as follows:

[tex]Pb(NO_{3})_{2}+2NH_{4}I\rightarrow PbI_{2}+2NH_{4}NO_{2}[/tex]

From the above reaction, it is clear that 1 mole of [tex]Pb(NO_{3})_{2}[/tex] reacts with 2 moles of [tex]NH_{4}I[/tex].

Molarity and volume of [tex]Pb(NO_{3})_{2}[/tex] is 0.680 M and 253 mL respectively.

Calculate number of moles of [tex]Pb(NO_{3})_{2}[/tex] as follows:

[tex]M=\frac{n}{V}[/tex]

Here, n is number of moles and V is volume. Convert volume from mL to L

Thus, [tex]V=253\times 10^{-3}L[/tex]

Rearranging equation and putting values,

[tex]n=M\times V=0.680 mol/L\times 253\times 10^{-3}L=0.172 mol[/tex]

Since, 1 mol of [tex]Pb(NO_{3})_{2}[/tex] reacts with 2 moles of [tex]NH_{4}I[/tex] thus, 0.172 mol of [tex]Pb(NO_{3})_{2}[/tex] will react with [tex]2\times 0.172 mol=0.344 mol[/tex] of [tex]NH_{4}I[/tex].

Now, molarity of [tex]NH_{4}I[/tex] is 0.290 M or mol/L thus, volume can be calculated as:

[tex]V=\frac{n}{M}=\frac{0.344 mol}{0.290 mol/L}=1.186 L[/tex]

Converting volume into mL

[tex]V=1.186 L\left ( \frac{1000 mL}{1 L} \right )=1186 mL[/tex]

Therefore, volume of 0.290 M [tex]NH_{4}I[/tex]  required will be 1186 mL.

The volume of a 0.290 m NH4I solution that is required to react with 253 ml of a 0.680 m Pb(NO3)2 solution is 1.186 liters or 1186 ml.

Further Explanation  

Chemical equation  

  • A chemical equation is an equation showing chemical symbols of reactants and those of products. They represent a chemical reactions between reactants to form products.
  • For example; Pb(NO3)2(aq) + 2NH4I(aq) →PbI2(s) + 2NH4NO3(aq) where Pb(NO3)2 and NH4I are reactants while PbI2 and NH4NO3 are products.  

Precipitation reaction

  • Precipitation reactions are reactions which involves the formation of a precipitate as one of the products. A precipitate is a compound that is insoluble in water.
  • Pb(NO3)2(aq) + 2NH4I(aq) →PbI2(s) + 2NH4NO3(aq) is an example of a precipitation reaction where the compound PbI2 is the precipitate.  

Moles, Molarity and Volume

  • Molarity or concentration is calculated by dividing the number of moles by volume in liters.
  • That is, M = n/V; where M is Molarity, n is the number of moles and V is the volume.

In this case our balanced equation is:

Pb(NO3)2(aq) + 2NH4I(aq) →PbI2(s) + 2NH4NO3

Step 1; Moles of  Pb(NO3)2

M = n/V

thus; n = M x V

Molarity = 0.680 M and Volume is 253 mL or 0.253 L

n = 0.680 x 0.253

   = 0.172 moles of  Pb(NO3)2

Step 2 Moles of NH4I

Mole ratio;

From the equation; 1 mole of  Pb(NO3)2 reacts with 2 moles of NH4I

Thus;

Moles of NH4I = moles of Pb(NO3)2 multiplied by 2

                         = 2 x 0.172 moles

                         = 0.344 moles of NH4I

Step 3 Volume of NH4I used

M = n/V

Therefore;

V= n/M

Molarity = 0.290 M and Moles = 0.344 moles

Thus; Volume = 0.344 moles/0.290 M

                       = 1.186 L

Therefore the volume of a 0.290 m NH4I solution that is required to react with 253 ml of a 0.680 m Pb(NO3)2 solution is 1.186 liters or 1186 ml.

Keywords: Molarity, moles, volume

Learn more about:

  • Relative atomic mass: brainly.com/question/4079113
  • Moles calculation: brainly.com/question/4079113
  • Molarity: https://brainly.com/question/12834126
  • Example of a question on molarity:https://brainly.com/question/12834126

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Molarity, volume and moles

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