Chemistry and TEAS 101

Question

What is the mass of the precipitate formed when 50ml of 12% solution of AgNO3 is mixed with 50 ml of 6.7% solution of NaCl?

Answer 1

Balanced equation: AgNO3(aq) + NaCl(aq) ==> AgCl(s) + NaNO3(aq)

We will convert AgNO3 and NaCl to moles and see if either is in limiting supply.

Next, we will find moles of AgCl (the precipitate) and then convert that to grams of AgCl.

AgNO3: 12% = 12 g/100 ml

12 g/100 ml x 50 ml = 6 g AgNO3

moles AgNO3 = 6 g x 1 mol/170 g = 0.0353 moles AgNO3

NaCl: 6.7% = 6.7 g/100 ml

6.7 g/100 ml x 50 ml = 3.35 g NaCl

moles NaCl = 3.35 g x 1 mol/58.5 g = 0.0573 moles NaCl

Since NaCl and AgNO3 react in a 1:1 mole ratio, AgNO3 is limiting

moles AgCl produced = 0.0353 mol AgNO3 x 1 mol AgCl/mol AgNO3 = 0.0353 moles AgCl produced

mass AgCl produced = 0.0353 moles x 143.5 g/mol = 5.07 g AgCl = 5 g AgCl (to 1 significant figure)

Answer 2

AgNO3+NaCl= AgCl+NaNO3 AgCl is precipitation, it does not dissolve in water

Let's find M AgNO3=170 NaCl=58.5 AgCl=143.5

50mL 12% solution of AgNO3 means 6 g of powder dissolved in 50 mL solution.

( 12% solution means - 12 gram of AgNO3 in 100 mL solution)

50 mL 6.7% NaCl contents 3.35 gram of salt NaCl

So, 6 g of AgNO3 mixed with 3.35 gram of NaCl. Let's find out if this is a right proportion.

look on their molar mass: 170 g AgNO3 will react with 58,5 g NaCl

As we can see that 3.35 g NaCl is too much to react with 6 g AgNO3.

We can find right mass of NaCl: (6 x 58.5) :170= 2.06 g

Since, 2.06 g is right number now, we can find precipitation AgCl

(2.06 x 143.5): 58.5= 5.05 gram of precipitation AgCl