Question

is al Gig H 40,- 32.2 mae Of c esent in a 30 moves 10/24111 11. Aluminum reacts with oxygen to produce aluminum oxide. Calculate the mass of aluminum oxide that can be produced if the reaction of 50.0 g of aluminum and sufficient oxygen has a 75.0% yield.

Answer 1

2008 & Solutions - Querit Given that:- mass of Aluminum (AI) = 50.09 Percent yield = 75.06 0 So Balance chemical Reactions - 4 Alcs). +30249) 4241203 ; molar mass of Ai = 27g/mo! So mole of Al EnAI) = 500g 27.0g/mol nal= 1,85 mol According to balance Reaction 4 mol Al Produce a mol ALLO3 i mol Al produce 2 mol Al2O3 1.85 mol Al produce 2x185 mol ALC3. mole of AbO3(MALO3) = 0.g26mol. molar mass of AlO3 = 102g/mol So Toertico yield of Al2O3 = 0.926 X 1024 molt 3.94. 4529 . Actual yiele x100 Thoertical yield put the value in the equation Actueel vjeld of A103 = 4 yield x Theertical yield 100 75.0 X 94.4528 100 mass of Aluminum oxide = 70:849 Ansis Que 2 2 Balance chemical Reaction: 50 + 2.502 → CS2 + 400 mass of carbon (c) = 40g mass of CS2 = 369 (Actual vield) A. Vielol Actual vielel

molar mass of carbon = 12g Imol so mole of Carbon (n.) = 40g - 12 g/mol . nc =3. 33 mol According to balance Reaction:- . 5 mol į Produce 1 mol csa. 1 mol c: Produce Imol cs₂ 3:33 mol c produce X3.33 mol CS2 mole of cs₂/ncsi) = 0.666 mol molar mass of ch₂ = 76.14 g/mol © Thuertical yield of cś2 = 0,666 mol X 76:14 g/mol => 50:71g viel xivo = 369 xloo => 70-9914 71% Ans Thoerticed yield ? 50.71g yied = Actueel vied in