Question

Part 1 (1 point) du See Periodic Table Sulfur trioxide dissolves in water, producing H2SO4. How much sulfuric acid can be produced from 11.3 mL of water (d= 1.00 g/mL) and 22.3 g of SO3? Part 2 (1 point) ♡ See Hint How much of the reagent in excess is left over?

Answer 1

Argas. SO₂(g) + H2O HasO 4 (aa) Molermass of 'SO3 32+ 3x16 = 80 g/mol Molurmass of H2SO4 = 2x1 +32+16X4 = 98 g/mol H20 = 2x1 +16 2 18g 1 mol Mass of H₂O= vxd 2 11.3 ml x 100g / me. = 11.3 g Part 1: 1 mol soz react with 1 mole H2O 80 dog so3 reacts with 18 g H2O so 22:39 363 will react with 18 Ex 2.38 420 = 5.028 H2O <11:3 s so H₂O is excess reagent ie soz is limiting reagent. 1 mole so gives 1 mole Hasu 4 dog soz gives 989 H2SO4 so 22.3 g 503 will will react with give _ 98 x 22.3 g H₂SO4 = 27.329 80 : mass of H2SO4 formed = 21:32a) part 2 Mass of excess reagent remain = 11:39–5:02g=[ 6.299] please sate it)