Question

i dont know what im doing. help.

4.3-Precipitation Reactions 3. The following molecular views show reactant solutions for a precipitation reaction (with H,O molecules omitted for clarity). a) Which compound is dissolved in beaker A: KCI, Na2SO4, MgBrz, or Ag2SO.? b) Which compound is dissolved in beaker B: NH,NO3, MgSO Ba(NO3), or CaFZ? c) Write a balanced molecular, total ionic, and net ionic equations for the reaction that occurs when solutions A and B are mixed. d) If 1 L of a 0.10 M solution of each reactant are mixed, what mass of precipitate can form (assuming a complete reaction)? 4.3-Precipitation Reactions 4. Magnesium is the second most abundant metal in sea water after sodium. The first step in its industrial extraction involves the reaction of Mg2+ with Ca(OH)2 to precipitate Mg(OH)2. What mass of Mg(OH), is formed when 0.180 L of 0.0155 M MgCl2 reacts with excess Ca(OH)2? 5. Iron(III) hydroxide, used to adsorb arsenic and heavy metals from contaminated soil and water, is produced by reacting aqueous solutions of iron(III) chloride and sodium hydroxide. What mass of iron(III) hydroxide is formed when 0.155 L of 0.250 Miron (III) chloride reacts with 0.215 L of 0.300 M sodium hydroxide?

Answer 1

А В a) The compound that is dissolved in beaker A is Na sou 6) The compound that is dissolved in beaker Bis Ba(NO) (c) Balanced chemical equation: Nasou + Ba(NO3) 2 Nang & Basolu (aq) (aq) (aq)

complete ionic equation :- 2Na agt sou cart Ba (a) * 2Na aaj + 2003 Go + Basou (3) Net ionic equation :- 2 + Ba + sou (29) (09) - Basou souls

Nem chemica UMCO0 d) from the balanced chemical equation, Imol. of Naso reacts with Imol. of Ba(NO), and form Imol. of Baso precipitate. Given 11. of o. lM of each reactant is reacting. That means, no. of moles - Molarity & volume in L. - 12.x 0.1M = 0,1 mol : 0.1mol. of Nasoy reacts with olmol. of Ba(NO3)2 It forms o.1 mol. of Basous

1 mol of Baso weighs (its molecules ) 233.4 g mass, 0.1 mol of Baso mass is 23.349 Hence, 23.39 of Basou precipitate is formed.

4. The chemical reaction is Mgent CaOH Mg(OH)₂ + cart (29) 209) (5) lap from this balanced chemical equation, Imol of Might reacts with 1 mol of CaloH), and form 1 mol of Mg(OH)₂ From the given data, Ca(OH) is excess in amount. Them, Macle or right becomes limiting reagent and the amount of precipitate forms depend on Mght only. Given, 0.1802 of 0.0155M of Mg2+

No. of moles - Molarity & Volume no. of moles of Might that is reacting is - 0.180L x 0.0155 M = 0.00279 mol. So, 0.00279 mol. of Mg2t forms 0.00279 mol of Mg(OH). Mass of 0.0027amol of Mg(OH)2 = ? Mass of 1 mol of Mg(OH)2 = 58.3 g. Mass of 0.00279mol. Mg6H), = 0.00279 mol. x 58.39 – 0-1629

5. The chemical reaction is: - Fecha cap + 3 Na OH (29) ► Fe (OH)3 (+ 3 Nad 199) from the above balanced chemical equation, it is cheas that 1 mol. Fech reacts with 3 mol of NaOH. Given, 0. 155 of 0.2509 Fech reacts with 0.2152 of 0.300M NaOH. =) (6. 1552 x 0.2504) Fech neacts with (0.215 L x 0.300M) NaOH = 0.03875 mol fech reacts with 0.0645 mol NaOH 1 mol. Fech reacts with 3mol. NaOH then 0.03875 mol. Fech requires how many moles of Nach 4 0.03875 mol Fech x 3 mol. NaOH 1 mol = feel 0.11625 mol NOOH

But only 0.0645 mol of NaOH is available So, NaOH is the limiting reagent which decides the amount of product Fe(OH). Feel, is excess reagent from the balanced chemical equation, 3mol. of NaOH forms – 1 mol of fel OH)₂ Then 0.0645 mol. NaoH forms how many moles of felot ? 0.0645 mol NaOH x 1 mol. Fe(OH)3 - 0.0215 mol. Fe(OH). 3 mol NaOH Mass of 0.0215 mol of Fe(OH = ? Mays of 1 mol of Fe(OH)3 = 106.869. => 0.0215 mol. x 106.868 - 2.299. 1 mol. Hence, 2.299 of FeloH), will be formed.