Question

Use the References to acce For the following reaction, 16.7 grams of chlorine gas are allowed to react with 50.5 grams of sodium iodide. chlorine (g) + sodium iodide (3) ->sodium chloride (s) + iodine (s) What is the maximum amount of sodium chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 8 more group attempts remaining

Answer 1

dus) give a) mass of chlorine gas = 16.79 moles of chlorine gas= ware y molare mass 16.79 70.9 g/mol = 0.236 mol b) mans of Na I = 50.59 males of NQI= 50.59 149.99 = 0.337 mol D Solution balance chemical Rxn. Cola (8) +2N0I(S) - 2 Nallis) + I2 (5) E stek ② find limiting reagent. E a) moles pere (officient of Cola= moles = 0.236 mol Cofficient b) molus pave cofficient of NOI = 0.337mol = 0.168 = 0.168 <0.236 so formula of Limiting Reagent = Na I.

Step II) Hind maximum mass of Sodium Chevice. applied PoAc rule on balanced Ron. cofficient moles of Nard - moles of Nai Cofficient mass of Mald = 0.33 Imal molau mats x z mass of Nach = 0.337 mal 58.4ug/mol mans of Nacd = 19.69 g dus applied again POAC rule mass of cela E moles of Nal malau mass & cofficient cofficlent mass of Cdo 0.367 70.9 XI mass of Cdy = 0.33 7X70.9 - 11.959 mads of Cele that hauticipate in Ren= 11.959 steh I find Remaining mass of Colg. Remaining mass of (de= Total mass - mass that React UUUUUUUUTTT Remaining mass of (da= 16.79 - 11.959 = 4.75g Thank you