Question

Name: 5) Sulfuric acid is one of the most important indu fertilizers. One reaction in the synthesis of sulfuric ac pts) at industrial chemicals and is used in the p is used in the production of acid ls summarized by the reaction below. 25 (s) +30, (g) + 2 H20 a. If 45 g of completely reacted, how man credit. (10 pts) (8) + 30, (g) + 2 H2O (1) + 2 H,50. (aq) etely reacted, how many grams of H,SO. would tom would form? Show all work for full b. If 55 g of Oz completely reacted, how many grams of H:50 credit. (10 pts) grams of H.SO, would form? Show all work for full c. If 45 g of S is combined with 55 g of O, and excess H20, the limiting reagent is . (2 pts) d. The theoretical yield of the reaction is of H2SO.. (2 pts)

Answer 1

given reaction is 25 (s) + 302(g) + 2 hores 2 t, sou cage In the given balanced equation we - see that 2 moles of sulphur (s) react with 3 moles of O and 2 moles of the to produce a moles of tesou. a) First we will calculate number of for which formuld to be used as: Number of roles - Mass Molar Mass As Molar Mass of s = 32 g/mol کیا سکه شعاع

Number of moles of s = Mass of s Molar mass of s - usg - 32 g/mol = 1.406 moles Since according to the balanced chemical equation 2 moles of s produce a moles of Hysense 1.406 molests will produce 1.406 moles of the sono Now using ③ we can calculate mass of Hy sou in 1.406 moles as Mass of the sou (No. of moles of sou) corolar mass of the sou) @ As Molar mass of the sou= 98.079 g/mol So, using we get Mass of the sou = (1.406 mot) & (98.079 g/mol) = (1.40 6 X 98.079) g 3137.99 ~138 g This 138 g of the sou would form if usg of completely reacted.

6 Giren mass of a 55g Molar Mass of a = 31.998 g) mol Using we get No. of moles of a sog - 31.999 imol = 1.719 moles According to the balanced chemical equation 3 moles of a produce thsou = 2 moles I mole of O₂ will produce te sou = 2 moles - 3 cofon will produce the sou2 2X1:219. 1.719 mo 3 1146 moles So Mass of the sou in 1.146 moles = (1.146 moles) x (98.019 g/mol) = 112.4g| Thus 112.4 grains of the sou would form f ssg of a completely reacted

c) No. of moles in us g of s=1.406 moles No. of selesin ssgopz=1719 moles According to balanced chemical equation 2 moles of s react with a = 3 moles mole of s will react with a = 3 moles So, 1.406 moles of will react معا للاحته حاجی - =2.109 moles Thus 1.406 moles (or usg) of s requires 2.109 moles of a but there are only 1.719 moles of a present in ssg of a Thus a is the limiting reagent. d) Theoretical yield is always calculated with respect to the limiting reagent as once the limiting reagent is completely consumed reaction will not be able to proceed. So 3 moles of a produce 2 moles of th504 1.719 moles of a produce 1.146 moles of the sou. As already calculated is parte)

I و And ۱ Mass of thou in 1.146 moles والله الها A This Theoretical field of the ها نحللن هنلممعع وبعد