25.00mL of 0.25M of AgNO3 react with 20.00mL of 0.150M of HCl. Find the mass of...

25.00mL of 0.25M of AgNO3 react with 20.00mL of 0.150M of HCl. Find the mass of precipitate.

Expert Solution

millimoles of AgNO3 = 25 x 0.25 / 1000 = 6.25 x 10^-3

millimoles of HCl = 20.0 x 0.150 / 1000 = 3 x 10^-3

AgNO3 + HCl ---------------> AgCl + HNO3

1 1 1

6.25x10^-3 3 x 10^-3

here limiting reagent is HCl . because mole ratio is low.

1 mol HCl -------------> 1 mol AgCl

3 x 10^-3 mol HCl -----------> ??

moles of AgCl = 3 x 10^-3 mol

mass of AgCl = 3 x 10^-3 x 143.32 = 0.430 g

mass of precipitate = 0.430 g

queen_honey_blossom answered 1 year ago

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