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In the laboratory 6.67 g of Sr(NO3)2 is dissolved in enough water to form 0.750 L. A 0.100 L sample is withdrawn from this stock solution and titrated with a 0.0460 M solution of Na3PO4. a. What is the concentration of the Sr(NO3)2stock solution? b. Write a balanced molecular equation for the titration reaction. c. How many milliliters of the Na3PO4 solution are required to precipitate all the Sr2+ ions in the 0.100 L sample? (MM's: Sr(NO3)2 = 211.64; Na3PO4 =163.94)Name

Respuesta :

Answer:

2

Explanation:

joseaaronlara

Answer:

The answer to your question is below

Explanation:

Data

mass of Sr(NO₃)₂ = 6.67 g       Final volume = 0.750 L

Sample  0.100 L

[Na₃PO₄] = 0.046 M

a) [Sr(NO₃)₂

MM = 211.64 g

                          211.64 g ---------------------- 1 mol

                             6.67 g ---------------------- x

                             x = (6.67 x 1) / 211.64

                             x = 0.032 mol

Molarity = 0.032 / 0.75

Molarity = 0.042

b)

             3Sr(NO₃)₂  +  2Na₃PO₄  ⇒  Sr₃(PO₄)₂  +  6NaNO₃

              Reactants             Elements         Products

                    3                           Sr                      3

                    6                            N                      6

                    6                            Na                    6

                    2                            P                       2

                  24                           O                     24

c)

Calculate the moles of Sr(NO₃)₂ in 100 ml or 0.1 L

Molarity = moles / volume

Moles = Molarity x volume

Moles = 0.042 x 0.1

Moles = 0.0042

                  3 moles of Sr(NO₃)₂ --------------- 2 moles of Na₃PO₄

                  0.0042 moles of Sr(NO₃)₂ -------- x

                  x = (0.0042 x 2) / 3

                  x = 0.0028 moles of Na₃PO₄

Molarity = moles / volume

Volume = moles / Molarity

Volume = 0.0028 / 0.046

Volume = 0.060 L or 60.9 mL

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