Question

I balanced the combustion reaction correctly but when it came to solving limiting reagant I got confused. Can you help with B-E

When C3Hs reacts with Oxygen gas, water and carbon dioxide are formed. A) Write the balanced equation for the reaction given above: (3 pts) 8G H 0,1850 1360 (3H₂ + 50 360 B) 1 220 grams of CHa react with 20 grams of Oz, how much co, can be formed? (3 pts) 360 my determine the limiting reactant, and write the answer in C) ( H.150 2 1 220g (3 to 1 molesy 24 te mos (Holes 25 Tangih l&H TM uug CO2 wg . lnd a 13 loe | 9.375 mke a / cang - 2 1 32 a 1so a tlimiting 125.442 = S. 9.375 molly c) What is the limiting reagent for the reaction in #7?_a v (3 pts) 20 .275 D) How much of the excess reactant is left over? (3 pts) a lot E) How many grams of water will be produced from this reaction? (3 des) 4H,0 4, е 20g he 24 , what is the F) If 20.0 grams of H,O are actually formed in the action described in problem percent yield of this reaction? (3 pts) 22 20

Answer 1

water and coucbon When CaHeactro with. Oxygen qan dioride are fonmed (A) balanesd equation forr lhe steaction, Catte 502 ->3Co 4H20 (Ans) 41 Molart mam of cgtq = (3X12)+ (8x1) mst /mo1= g/ma I| Molorr mam of o2= (2x16)g/ma 32 g/mol of o, 220 gm af c3 Hg eact with 20 220 9 of 3g 22s m C3g 41 almo 5 mol of Ca H 320 32 g/mel mol of 02 o 2.5 Respect to CaH 1 mol of CaH producen 4 mol of H20 of erg 5 mo A x5 molaf imelaflgs 20 maof Ho 20 mol af H20 20 X \) of 20 360 of H20 Tmclar mam of Ho = 18 glmol Renpect to o2 pocoduca 4 mol of H20 Б mol 4 mol of H20 x o625 malof o 5 molaf 2 Praducan of 0.625 mo) a.5 mal of Hzo o.5 mol af H20-- (d-5x 18 of H20 so, 2 in the imiting gent Now mol of o producar 3 mol of co 3 mol of C02 x o. G25 matof on O.625 mol of on 5 molafez o 375 mol of co molar mam of co2 = 44 g/mol 0.35 mol ot Co, z (o.375 x 4 4) = 16 5 of coL

165 am ef co2 can be formed, k ol0 (Ans) , (oxyaen) in the limi ting zeactant for the Jearion (4ns) 1 mol of C Hg Now 5 moi of 0, react with moi of g Hg X 0-<25 mator o, 5 molo2 0.625 mo f o, iaatt with 125 mol of aHg- o125 mol of cgrlg - 0.125 mol x 4g/mol) of (aH- (220-5 5.5) 214.5 gram of exam oieackant (caH8in ieft over. (Ans) 4 mal of 2 ptoducan mal 1mo of H20 x o625 mglof B .625 5 mol af .5 mol of H20 = (0-5 mol x 6 q!moi) of H20 of H2owalen) will ke priadu co.d from the eartion. (Ans) (F) Experimentai yield = 20 gm of H 0 Fxperimental Yield thamritical Yield Theoritica Yield -X 100 Yield percend ( y.) = 9 x 100 Y. 20 45 percent yield of the eaction in 45. (Ans)