Question

For the following reaction, 0.362 moles of sulfur dioxide are mixed with 0.328 moles of water sulfur dioxide(g) + water() - sulfurous acid (H2SO3)(g) What is the formula for the limiting reagent? What is the maximum amount of sulfurous acid (H2SO3) that can be produced? moles

Answer 1

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50,(8) + Hgoll) → Hys03 (3) - Hq 0.362 mol 0.328 mol H al- = 23.c = "9 from stochiometry molar ratio of so, and H90 és 1:1 con = [H . [ Heo is limiting & reagent o Moles of 3 H₂SO₃ produced oo.si = 49 = -moles of H₂O - cost : Maximum amount of H₂SO3 produced - 0•328 moles ( 2 – 50 - 1