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A silver nitrate solution contains 14.77 g of primary standard AGNO3 ( Molecular weight 169.87) in 1.00 L. What volume of this solution will be needed to react with 0.2631 g of NaCl ( Molecular weight 58.44) ?
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Respuesta :

sebassandin

Answer:

[tex]V=5.2 mL=0.052L[/tex]

Explanation:

Hello!

In this case, since the chemical reaction between silver nitrate and sodium chloride is:

[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]

We can see there is a 1:1 mole ratio between each solution; thus, we first compute the moles of each reactant considering their molar masses:

[tex]n_{AgNO_3}=14.77g*\frac{1mol}{169.87g}=0.087molAgNO_3\\\\ n_{NaCl}=0.2631g*\frac{1mol}{58.44}=0.0045molNaCl[/tex]

Now, since the concentration of the silver chloride solution is 0.087 M, we may assume that the concentration of the NaCl solution is the same, so we can compute the volume as shown below:

[tex]V=\frac{n_{NaCl}}{M}=\frac{0.0045mol}{0.087mol/L}\\\\V=0.052L[/tex]

Or:

[tex]V=5.2 mL[/tex]

Best regards!

The volume of solution needed to react with 0.2631 g of NaCl  is 0.052 L.

How we calculate the volume?

Volume of the solution will be calculated by using the below formula:

M = n/V, where

M = concentration in terms of molarity

n = no. of moles

V = volume

Given chemical reaction is:

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

First we calculate the moles of given reactants by using the formula:

n = W/M , where

W = given mass

M = molar mass

Moles of AgNO₃ = 14.77g / 169.87g/mole = 0.087 mole

Moles of NaCl = 0.2631g / 58.44g/mole = 0.0045 mole

Concentration of AgNO₃ = 0.087 mole / 1L = 0.087M

From the stoichiometry of the reaction it is clear that mole ration of AgNO₃ & NaCl is 1:1. So, we take the concentration of NaCl is equal to the concentration of AgNO₃ and calculate the volume by using the above formula as:

Volume of NaCl = 0.0045mole / 0.087M = 0.052 L

Hence, 0.052 L is the required volume of NaCl.

To know more about moles, visit the below link:

https://brainly.com/question/17199947

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