Question

7. Given the following equation: 2 NaClO3 2 2 NaCl + 3 02, if 12.00 moles of NaClO3 reacted in the reaction; a) How many grams of O2 produced? B) How many grams of NaCl are produced when 80.0 grams of O2 are produced? 8. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2AI + 3Cl2 → 2AlCl3 a) Identify limiting reagent b) How many grams of aluminum chloride could be produced from 34.0g of aluminum and 39.0 g of chlorine gas? c) what mass of excess reagent remains in the reaction?

Answer 1

2 Nacioz 2 Nac + 302 o in 2 12 mole mole Nacioz gives 3 more Nalloz will give 02 3x12 mole O2 = 18 mole O. mass in gm = no. of moles x molar mass. à Grams of a produced = no. of moles of or produced & molar mass of O. = 18x32 g 02 = 576 g 02 © 80 g 02 =- 80g 32 g/mole = 205 mole Oz No. of moles = given mass molar mass from the reaction stoichiometry we can say that - o produced means 2 mole Nail also produced - 2.5 mole O₂ produced means 5) mole Nall -1.67 mole Naul produced