Question

Octasulfur(Sg) is the most common allotrope of sulfur and is widely used in the chemical industry, especially in the production of petroleum. While Sg is the main component of elemental sulfur on earth is can also be produced in the laboratory from the reaction of sulfur dioxide with hydrogen sulfide. The balanced chemical equation for this reaction is shown below: 8SO2(g)16H25(g)3S(s)+16H2O (g) Consider the reaction of 93.0 grams of SO2 with 93.0 grams of H2S to answer the questions below. Premise Drag and drop to match Response 140 grams What is the limiting reagent? A 8 HS What is the excess reagent? 2 =c87.1 grams 3 How many grams of Sg can be formed? How many grams of excess reagent are left over? =D 18.5 grams 4 E Fullscreen 个

Answer 1

Following evention is - 850,(g) + 16 H2SCQ) -> 358 ) +1672069) M.M= 8x 64.06 M.M = 16x 34.1 mm = 3x256 M.M = 16x18 M.M = 51 2.48gluule. M.M= 545 balul M.M=768glul mim = 288 plural mass = 93.0 і мол ) Cginen) mass = 93.0 oo M.M= Molar mass of the given reautant & procent. How we are eveantant & we know going to calculate the makes of the following product. 1 mole = Mass given). I Molar maust role = 93 = 0.18 141 S12:48 5456 8 SO2G) + 16H25Cg) 35868) +1642069) male = 0.1814. mole = 0.17045 mode = 93.480-1814) mole = aus. 650-17045) so the reartant which have less no. of mober is known as limiting neagent. . Because reaution is govern by the no. of moler. Hene H₂S is the limiting reagent. (because o 17045 mole). less mo. of musles. D 00 a reagens ♡ Mes is the limiting reagiert. - have more moles is the encen The reartent which 91eagen reagent.

so we can say by looking at the moler S02 is the excess reapest 3) How many grams of Sp is formed 850, (g) + 16 2.8.cg) -> 35&(8) +1622099). to 0-1814 nole 0.17045 hole. tzt 0.1814 -0.17045 0-17045 mole left = 0.01095 so 358 have we know 0.17045 mole. (govern by no. of males of 2.R limiting moler mass of 388 = 768 gluale. mole = mass molar mass 0.17045 = Mass 768 mars = 130.9 g a 131g. 3) 131g of S8 is formed. 4) soz us in emers. P 850₂ C) + 16 H₂S (9) s 358 (s) + 1642009) t=0 0.1814 mole 0.170 45 wale t= t 0.1814-0.17045 - o.ltorsnele o.17045 mole left - 10.0109Snole. o this is the amount left. we knew MiM of 8502 - 512.48 quele.

mole mars M.M oooloas = mass 512.48 mass = 5.611 ~ 5.77 to (1) limiting neagent - (2) Emers reagent =) (3) S8 formed =) (4) emers neagent Left = over B (H₂S) H (202) EC131g). f (5.77g).