Question

b. How would you prepare 225.0 mL of 1.33 M HCl from a 6.00 M stock solution? C. When 25.00 mL of 0.695 M HCl reacts with an excess of silver nitrate, will a precipitate form? Write the net ionic equation for this reaction. How many grams of precipitate can be theoretically obtained? d. What volume of 0.2500 M strontium hydroxide is required to completely react with 75.00 mL of 0.07942 M HCI? e. When 37.5 mL of 0.439 M HCl reacts with 22.0 mL of 0.573 M ammonia, ammonium ions are formed. What is the concentration of each species in solution after the reaction is complete? (Assume the volumes are additive.) f. An alloy containing aluminum is analyzed. All the aluminum reacts with 212 mL of 0.493 M HCl in the 2.500 g sample of alloy. Calculate the mass percent of aluminum in the alloy. (Assume that the only component of the alloy that reacts with HCl is aluminum, and that the products of reaction are hydrogen gas and aluminum chloride.) 36. Which of the following aqueous solutions has the greatest total concentration of ions present? a. 0.01 M HCI b. 0.05 M Cabr, C. 0.02 M Ca(NO3)2 d. 0.02 M Al(NO3)2 37. Which of the following metal ions form sulfate compounds that are insoluble in water: (2) Ba? (3) Ag* (4) Mg²+ (5) Pb ?? (1) K* a. 1 and 2 b. 3 and 4 C. 1 and 3 d. 2, 3, and 5 e. 3, 4, and 5

Answer 1

e) Both HCl and NH₃ are strong. So one equivalent of each reacts to neutralise.

b. is need to prepare let, a mit 225.0 ml 6.om soek soration of 1.33 on Hel. 2 x 6 x = => = 225 * 1.33 2.25 1.33 = 49.875 ml [ 49.875 ml Hence, day taking 49.875 mt, 6.om hel stock solution and assing (2.25-49.875) m = 175.125 ml of the info it for the preparation of 225 m nel. wy Ter, precipitate win form and enactly 25.0 me of 0.695 m precipitate form e The nef ionic equation is given below - Ag (+ g) + ele (an) Agelen Now, 25.0ml of o.695M Age = 1435 * 25 * 0.695 1000 g Ages = 2.493 g Agel. 2.493 g of Azes Precipitate will form theoretically SoH) is a diacidic berse ie if gives two hydroxyl groups pon moleme Son Concentration of 8 ion will be (0.25 x 20 m - 0.50m. of Again, 75 ml 0.07942 M Hel & 76 0.07942 0.5 11.91352 of osmael one egnivalent of son s necessary to neutralize one equivalent of Her Again each sne produce two equivalent d. 11 9 13 mt of of mr. Hey = 0:25 M 11.913 me. 6.56** mL Dzen sof? Sn (0) will de necunom 1.913 mt of 0.25m neupreliso y to mestno

Now, 22.0 mL of 0.573 M NH₃ is equivalent to

(22 * 0.573)/0.439 = 28.715 mL of 0.439 M NH₃

Therefore, after the completion of the reaction ammonia will be fully protonated and concentration of ammonium chloride will be 28.715 mL of 0.439 (M)

Concentration of HCl will be (37.5 - 28.715) = 8.785 mL of 0.439 (M)