When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation? AgNO3(aq)+KCl(aq)→AgCl(s)+KNO3(aq)

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Since you’ve got the first part, I’ll just answer the second question, so to speak.

So, you’ve got your balanced chemical equation for this reaction

##AgNO_(3(aq)) + KCl_((aq)) -> AgCl_((s)) + KNO_(3(aq))##

You know that you have a ##”1:1″## between all the species involved. This means that the number of moles of potassium chloride must be equal to the number of moles of silver nitrate.

Since I assume you’ve calculated the number of moles of silver nitrate to be

##C = n/V => n = C * V##

##n_(AgNO_3) = “0.162 M” * “1.27 L” = “0.206 moles”## ##AgNO_3##

automatically you’ll have

##”0.206″cancel(“moles “AgNO_3) * (“1 mole” KCl)/(cancel(“1 mole “AgNO_3)) = “0.206 moles”## ##KCl##

Now use the volume of potassium chloride given to figure out what the solution’s must be in order to have that many moles of ##KCl## available for the reaction

##C = n/V = “0.206 moles KCl”/”3.78 L” = color(red)(“0.0545 M”)##

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When solutions of silver nitrate and potassium chloride are mixed, silver chloride precipitates out of solution according to the equation? AgNO3(aq)+KCl(aq)→AgCl(s)+KNO3(aq)

Leave a Comment / Uncategorized

Since you’ve got the first part, I’ll just answer the second question, so to speak.

So, you’ve got your balanced chemical equation for this reaction

##AgNO_(3(aq)) + KCl_((aq)) -> AgCl_((s)) + KNO_(3(aq))##

You know that you have a ##”1:1″## between all the species involved. This means that the number of moles of potassium chloride must be equal to the number of moles of silver nitrate.

Since I assume you’ve calculated the number of moles of silver nitrate to be

##C = n/V => n = C * V##

##n_(AgNO_3) = “0.162 M” * “1.27 L” = “0.206 moles”## ##AgNO_3##

automatically you’ll have

##”0.206″cancel(“moles “AgNO_3) * (“1 mole” KCl)/(cancel(“1 mole “AgNO_3)) = “0.206 moles”## ##KCl##

Now use the volume of potassium chloride given to figure out what the solution’s must be in order to have that many moles of ##KCl## available for the reaction

##C = n/V = “0.206 moles KCl”/”3.78 L” = color(red)(“0.0545 M”)##

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