Question

T14HW Question 13 Homework . Unanswered Fill in the Blanks A solution containing 58.0 g of mercury(II) nitrate is allowed to react completely with a solution containing 16,642 g of sodiur sulfate according to the equation below: Hg(NO3)2(aq) + Na2SO4(aq) + 2NaNO3(aq) + HgSO4(s) How many grams of solid precipitate will be formed if the reaction has a 100% yield? Solid precipitate = 1 grams How many grams of the excess reagent will remain after the reaction? Excess reagent remaining = grams If the reaction has only an 80% yield, how many grams of solid precipitate will be formed? Solid precipitate 2 grams

Answer 1

To ve Cuvon Mare C Molas Mars - 1429 Son Civen reach Mg(NO3)2 (aq) + N92502100) --2NaNoz (29]+ Mg 594 ($ fluu is the leelanced reach . Precipitate Mass of Kg (NO3)₂ = 589 A mole of Hg(NO3)2 = Given Mary S8g =001786ml Mola Mass 3248 /mal Кам ма, со = 16-сча 16.6429 - Mlale q Naz Soy - Creme de sa juzs/mob 01/17.2 miel So Na, sou ei limiting reeagent] → for 100% yield - Imole of Na, sou give Imole of Hg Soyls) 8. 1192 male of No,sou give 0.1122. mal of Hg so,(s) Mars of Hg souls) != Mugiso, Meggoy 0.1172 X 296.41 * Excess fecegent offer secara oven mole of log (Noz) = 0.1986-0192 20.0614 nell • Mass of Hglnogh left z o.oblumdX 324,8 g/mol pog] . %yield> Boy wield - Albeul omale ximo 3 r. yield sete fluorell Mars 80 = Actueel mas xico SE34.88 Actual Mars of regsau = 27.849) may = 34.8 nelx 324 88/mal