Question

Hydrogen sulfide reacts with oxygen to give sulfur dioxide and water according to the following balanced equation: 2 H2S (g) + 3 O2(g) 2 SO2(g) + 2 H20 (8) If we allow 8.52 g of hydrogen sulfide (H2S) to react with an ex a) Calculate the number of moles of SO2 that can be produced from b) Calculate the mass in grams of SO2 that can be produced from znat can be produced from this reaction: ac c) If the mass of SO2 produces is actually 11.2 g, what is the percent yiel Show your work, use correct units and significant figures in your results

please show work

Answer 1

2 Has (g) + 302 - 2802(g) + 2 H₂O(g) mass of Has = 8.52gm. molar mass of his = 34 g/mol moles of Hes= 8.52gm = 0.25 moles 3491ml cow given that of is in Excess, hence Has is over limiting - Reagent calculation of product formation done using H₂ from Reaction 2 moles Has produced - 2 molesso2 1 mole Has produce » 1 mole so2 0.25 mole Has produce = 0.25 moleso2 molar mass of so2= 64 g/mole moles of 2.502 produce = 0.25 mass of son produce = 64*0.25 = 16 g/mol 10 - Actual mass of so 2 = 11.2gm. percentage yield actual mass theoritical mas o = 11. 29m Al 169m 40%