# Chenistry

A 0.8870g sample of a mixture of nacl and kcl is dissolved in water and the solution is then treated with an excess of agn03 to yield 1.913g of agcl. Calculate the percentage by mass of nacl in the mixture

1.913g/ 143 = 0.0133 M agcl
x/58 + Y/74 (0.887g-x) = 0.0133

Who helps me for your solution clearly step by step.

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1. hi! in order to solve this you will need to incorporate a bit of algebra. first off, you have to produce the equations for each reactions.

NaCl + AgNO3 ---> AgCl + NaNO3
KCl + AgNO3 ---> AgCl + NaNO3

and let NaCl = x ; KCl = y

since the sample is equal to 0.887, and your sample contains only NaCl and KCl, then we have our first equation as:

x + y = 0.887 ----- (1)

and our precipitate is equal to 1.913. In order to complete the equation for the precipitate, we will relate NaCl to AgCl and KCl to AgCl by:

AgCl from NaCl:
x milligrams NaCl * (Mol wt AgCl/Mol wt NaCl)
x mg NaCl * (143.32 mg AgCl/58.44 mg NaCl) = 2.45x

and AgCl from KCl:
y milligrams KCl * (Mol wt AgCl/Mol wt KCl)
y mg NaCl * (143.32 mg AgCl/74.55 mg KCl) = 1.92y

with this, our second equation for the precipitate will be:
2.45x + 1.92y = 1.913 ---- (2)

Now, with our first and second equation, with two equations and 2 unknowns, input in your calculator and you will get an answer equal to:

x (NaCl) = 0.3962 mg
y (KCl) = 0.4908 mg

Now we can get the percent NaCl:

%NaCl = (0.3962/0.887)*100

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posted by iza m.

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