[FerndaleTiger]

1) How many grams of precipitate are formed if 192 mL of a 0.889 M aluminum sulfate solution and 346mL of a 4.93 M sodium hydroxide solution are mixed together?

The balanced equation i found:

Al₂(SO₄)₃ + 6 NaOH ---> 2 Al(OH)₃ + 3 Na₂SO₄

So then I took .889 M x .192 L = .170688 mol aluminum sulfate then multiply by 2 mol Al(OH)₃ over 1 mol Al₂(SO₄)₃ multiply once more by 77.98 g Al(OH)₃ over 1 mol of that = 26.6 g Al(OH)₃.

The other part was 4.93 M x .346 L = 1.70578 mol Sodium hydroxide, multiply by 2 mol Al(OH)₃ over 6 mol NaOH multiply again by 77.98 g Al(OH)₃ over 1 mol of that = 44.4 g Al(OH)₃.

Now is this a limiting reagent or something? If so I'd choose 26.6 g as the answer...Help would be appreciated.

2) You are asked to prepare 0.162 L of a solution that is 0.612 M in acetate ion. Your only source of acetate ion is a bottle of calcium acetate solution marked 1.90 M calcium acetate. What volume (mL) of the calcium acetate solution must be used.

Do i find the moles of calcium acetate by the moles = Molarity/Liters and then work my way to to something? I confused.

*Edit: The Dilution problem was setup to use the formula Mi x Vi = Mf x Vf and then convert to mL and divide by 2. I understand what I did wrong for #2.

[AWK]

1. OK
2.  This formula works for acetate concentration, so double concentration of calcium acetate and result should be 26.1 mL.