Question

89. For each of the following unbalanced chemical equations, suppose 25.0 g of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface. a. CH3OH(1) + O2(g) + CO2(g) + H2O(1) b. N2(g) + O2(g) → NO(g) c. NaClO2(aq) + Cl2(g) → C102(g) + NaCl(aq) d. H2(g) + N2(g) → NH3(g)

Answer 1

divide their dh ordei to find Limiting yeagend /reactant, we num bei ot moles with thei stoi chiometic coefbicient repec tively Whase ever Value will be lowed will be dimi-ling reagent/reactant. 1eaction wilproceed according it whic h means product And will form acording to it

C2 HDHU) 3023) 2 CO2 () 3 tbOu) a) mas of CHrOH 2Sg Mw f CHsOH- 46 9/mal Cniven mas So, mol o CHOH Mw 259 46 8/mol 0.5435 mol mass O2 25g 259 mol o 0 32 g/mol 0. 72וצ mo for, CtOH O435 0r5425 for O 78125 O 26042 3 So, 0, has ow value it is the imiting reactant. Product wil) fom according to i now, 3 mol 0, reacls to produce 2 mel CO to produce mel Co o 02 veacts L mol mol ca 0.781 25 mol o0 reacts to produce 2xo7125 S mol o co 2 poduced- 0.S20P mol So, mas co pvodued mal x Mo 0.5 203 molx 44 8/mol 22.929

N2 8) 2N) 02 mas of N2- 259 mol of N2 259 289/mol 0.893 mol mas 02 25 g mol of Or 25 323/mal - 0.7812 me So, for N2 0 893 O 78125 Oת2ונרי Se, it is the limiting reactant and product 02 has dow Valu will foim accolding to t 2 mol N0 i mol e 0 po cuces now 0 78125 mol 0 procduces 2GI 125 mol NO So mol NO poduced 625 mol ma No broduceel mol x Mo S625 mal 308/mol 46.875 9 2Nac car 2C102 (3) C8) c) 2Nauora) mas oNac02 25g Mw of Na uo,- 90.459/mol 259 2 0 2764 mol SO, mol o Na C0, 259 mas of mole o C 259 70 93/ma O 3526 mal 0-2744 for Nacio, O1382 O.1526 0.152

Nacio has low value So, Naclo, isthe limiting reactand Pro duct will form according to t nous 2 mol of Nacio, poduces 6Nado pacrces 2 mol ef Nac 1 mol of Nacd 1 mol O 2764 mal f Naco, broduces 0276 mol b Nact So, mol ofNa Cd procked 6.2764 mol mol x Mu So, mau Nac broduced 02764 malx S8.4m -16-1563 2NH3 g) N2 ) d) 1tl2 0) mass H2 25g mol of He 259 2 3/mol 12 S mol mau dN2 mol N2 25g 25 219/mo 12-Smet 4.167 0 892 mol foi H2 o.893 -0P93 for N2 Value So, N2 is the limiting reactand Pecuct N2 has the ave will torm according to r 2 mol of NH now, Produces 1mol N2 Pradures 2x083 mol of N .0.893 mol N2 So, mol ofNH Prodeced omas of NH3 brocu ced mol x Ms 178 6mol 726 mol x 179lma 30.3629