Question

For the following reaction. 39.5 grams of sulfuric acid are allowed to react with 26.9 grams of calcium hydroxide. sulfuric acid (aq) + calcium hydroxide) calcium sulfate (s) + water (1) What is the maximum amount of calcium sulfate that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

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Answer 1

The balance reaction Calcium of sulfuric acid and hydrocide is as followers. H2SO4 (aq) + Ca(OH)2 (5) → Caso4(99 + 2H200-0 Melar mass of sulfuric acid = 98.079 g/mol Molas mass of- calcium hydroxide = 74.093/mol Co 39.5 grams of Sulfurie acido = 39.5 moles of sulfurie neid 98.079 ? 0.40274 moles of Suefurie acid 269 grams of calcium hydroxide = 26.9 moles of calcium hydrovide 74.093 - 0.36306 moles of calcium hydroxide From equations, we observed that 1 mole H₂SO4 react with 1 mole Ca(OH)2 I to form 1 mole Casoy (Calcrum Sulfate) So 0.36306 mores of CaOH react with 0.36306 moles of Hy sou to give 0.3 of calcium Sulfate Molar mass of calcium sulfate = 13.6•149/mo So 0.36306 moles Cas04 = 49.427 gxames of Caloy So maximum amount of formed Casou = 149427grans 7 From the given data we can observed that the amount of calcium hydroxide, is lowey than that of required to reach with Sulfurie acid . So Calcium hydroxide is. limiting reagent. The formula of limiting reagent is [Ca (OH)2 ] les of Hy 504 to give 0.36306 moles

7 we observed that 0.36306 moles of-Ca(OH), react with 0.3 6306 moles of - H2504. So the excess reg reagent be. the excess #2504 = (0.40274 - 0.36306) moles of H2SO4 = 0.03968. moles of H2S04 = 0.03968x 98.079) grams of H2SO4 = 3.892 grams of H2504. So the excess reagent = 3.8927 grams