When the gases dihydrogen sulfide and oxygen react , they form the gase sulfur d

Question

When the gases dihydrogen sulfide and oxygen react , they form the gase sulfur dioxide and water vapor. a) write the balanced equation for the reaction. b) How many grams of oxygen are required to react with 2.50 g of dihydrogen sulfide? c) How many grams of sulfur dioxide can be produced when 38.5 g of oxygen reacts? d) How many grams of oxygen are required to produce 55.8 g of water vapor?

Explanation / Answer

a) H2S(g) + (3/2)O2(g) ==> SO2(g) + H2O(g) b) moles of H2S = 2.50/34.0 = 0.073 Thus, moles of O2 required = (3/2) x 0.073 = 0.11 thus, mass of O2 = 3.52 g c) moles of O2 = 38.5/32 = 1.203 Thus, moles of SO2 produced = (2/3) x 1.203 = 0.802 Thus, mass of SO2 = 51.33 g d) moles of H2O = 55.8/18 = 3.1 Thus, moles of O2 required = (3/2) x 3.1 = 4.65 thus, mass of O2 = 148.80 g

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