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When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of the solution according to the equation:
AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq)

mass of AgCl =46.6 g

The reaction described in Part A required 3.95L of sodium chloride. What is the concentration of this sodium chloride solution?

When solutions of silver nitrate and sodium chloride are mixed silver chloride precipitates out of the solution according to the equation AgNO3aqNaClaqAgClsNaNO class=

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The concentration of the sodium chloride would be 0.082 M

Stoichiometric calculations

From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M

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The concentration of sodium chloride would be 0.082 M.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Stoichiometric calculations

From the equation of the reaction,

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

The ratio of AgCl produced to NaCl required is 1:1.

Mole of 46.6 g AgCl produced = [tex]\frac{46.6}{43.32}[/tex] = 0.325 moles

Equivalent mole of NaCl = 0.325 moles.

Molarity of 0.325 moles, 3.95 L NaCl

Molarity = [tex]\frac{Moles \;solute}{Volume}[/tex]

Molarity =[tex]\frac{0.325}{3.95}[/tex]

Molarity = 0.082 M

More on stoichiometric calculations can be found here:

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