2H2S+SO2--->3S+2H2O. When 7.50 g of H2S reacts with 12.75 g of SO2 which statement applies?
The coefficients in a balanced chemical equation determine the ratio of moles of reactants to moles of products. In this reaction, two moles of hydrogen sulfide reacts with one mole of sulfur dioxide to produce three moles of sulfur and two moles of water. This information is used to determine if there is an excess amount of one the reactants. To determine the number of moles, divide the mass by the mass of one mole. For H2S, mass of one mole = 2 + 32.1 = 34.1 grams n = 7.50 ÷ 34.1 This is approximately 0.22 mole. For SO2, mass of one mole = 32.1 + 2 * 16 = 64.1 grams n = 12.75 ÷ 64.1 This is approximately 0.2 mole. According to the coefficients in a balanced chemical balanced equation, the ratio of moles of H2S to moles of SO2 must be 2:1. In this problem, there is excess SO2. This means all of H2S reacts. According to the coefficients in a balanced chemical balanced equation, the ratio of moles of H2S to moles of S is 2:3. Moles of S = 1.5 * 7.50 ÷ 34.1 = 11.25 ÷ 34.1 This is approximately 0.33 mole. Mass of sulfur = 32.1 * (11.25 ÷ 34.1) = 361.125 ÷ 34.1 This is approximately 10.6 grams. So b is the correct answer. I hope this helps you to understand how to solve this type of problem.
SO2 reacts with H2S as follows... Multiple choice.?
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SO2 reacts with H2S as follows: 2H2S + SO2 → 3S + 2H2O When 12.5 g of H2S reacts with 5.65 g of SO2, which statement applies?
12.5 g H2S x 1mole H2S/34.0g x 3 moles S/2moles H2S = 0.551 moles S 5.65 g SO2 x 1mole SO2/64g SO2 x 3 moles S/1 mole SO2 = 0.264 moles S One cannot obtain more than 0.264 moles S. The limiting reactant is SO2. 0.264 moles S x 32.1 g S/1mole S = 8.47 g S
SO2 reacts with H2S as follows: 2H2S + SO2 ----> 3S + 2H2O 7.50 g of H2S reacts w/ 12.75 g of SO2, which statement is correct?
A. 6.38 g of Sulfur are formed B. 10.6 g of sulfur are formed C. 0.0216 moles of H2S remain D. 1.13 g of H2S remain E. SO2 is the limiting reagent I found out the moles of H2S (0.110) and SO2 (0.199). This cancels out choices (C), (D) and (E). Now that I know H2S is the limiting reagent, I multiply it's moles by 3/2 to get the moles of S (0.165) and then multiply that by the molar mass of sulfur to get the mass of how much sulfur is formed. I keep getting 5.28 g of sulfur is formed and that isn't that correct answer, I know (B) is. What am I doing wrong?
Looking at the following information, which is the limiting reagent?
The reactants are of direction HCl and Ca(OH)2; we additionally see in accordance to the coefficients that we want two times as a lot HCl as Ca(OH)2. If we've a million.4 moles of Ca(OH)2 we are going to wish 2.8-moles of HCl. provided that we've greater desirable than that (we've 3.5 moles), Ca(OH)2 is the proscribing reactant. I used to apply bologna and cheese sandwiches to instruct proscribing reactants: enable's say a sandwich includes 2 slices of bread, a million slice of bologna, and a pair of slices of cheese. If we've 9 slices of bread, 3 slices of bologna, and eight slices of cheese, what share sandwiches will we make? 9 breads = 4 sandwiches, 3 bolognas = 3 sandwiches, 8 cheeses = 4 sandwiches. The bologna limits the form of sandwiches...we are able to in effortless terms make 3. we are able to of direction have bread and cheese left over!
Potassium permanganate is a very strong oxidising agent (stronger than sulphuric acid). In acidic medium, Mn[math]Mn[/math] gets reduced from its +7 oxidation state to +2 oxidation state. (n-factor is 5).2KMnO4+3H2SO4⟶K2SO4+2MnSO4+3H2O+5[O]Since SO2 is a strong reducing agent, it reduces Mn[math]Mn[/math] as denoted by the equation5SO2+2KMnO4+2H2O⟶2H2SO4+2MnSO4+K2SO4Both the reactions are viable. Because of the formation of MnSO4 and the disappearance of KMnO4, decolorisation occurs.But sulphur dioxide reacts with water to produce sulphurous acid and not sulphuric acid.SO2+H2O⟶H2SO3Hence, the acidic nature reasoning doesn’t hold goodthnxx for asking ……..and don’t forget to upvot.for latest update for current affairs and GK plz visit >>>>>suboyblog.
This reaction is taken as an experimental verification for the presence of sulphur dioxide gas (SO2).The equation for the reaction may be stated as follows:-K2Cr2O7 + H2SO4 + 3SO2 ——— K2SO4 + Cr2(SO4)3 + H2O.The orange-coloured dichromate solution will turn green due to the formation of chromium(III) sulphate, Cr2(SO4)3.It is a redox reaction. Here:-The oxidising agent is potassium dichromate and the reducing agent is sulphur dioxide.Hope it helps!!!!
This is a simple question of Limiting Reagent.Here is the balanced Equation :2H2S +SO2 = 2H2O + 3SThus if 2 moles of H2S reacts with 1 mol of so2 then 3 moles of S will be formedBut in the given condition, SO2 is only half mole (11.2L of gas at STP represents half mole), thus it is limiting Reagent.Thus if 1 mol of SO2 gives 3 moles of sulfur, then 0.5 Moles of it will give 1.5 moles of sulfur
Limiting Reagent Problem?
2 H2S + SO2 → 3 S + 2 H2O (7.50 g H2S) / (34.0814 g H2S/mol) = 0.220061 mol H2S (12.75 g SO2) / ( 64.0644 g SO2/mol) = 0.199018 mol SO2 0.199018 mole of SO2 would react completely with 0.199018 x (2/1) = 0.398036 mole of H2S but there is not that much H2S present, so H2S is the limiting reactant. (So not answers C, D or E.) (0.220061 mol H2S) x (3/2) x (32.0655 g S/mol) = 10.6 g S So answer B.
Hydrogen sulfide ( H2S) reacts with Sulphur Dioxide ( SO2) to form Water(H2O) and solid Sulphur (S).H2S + SO2 —> H2O + S(solid)Now as per the law of conservation of mass, mass is neither created nor destroyed in a chemical reaction. So, the equation should be balanced.Balancing the equation we have2H2S + SO2 = 2H2O + 3S(solid)I hope you get your answer. For further details or doubts, comments are welcomed in the comment box.[ AJ ^_^ ]