10.0 grams of Al and 19.0 grams of O2? Reaction: 4Al + 3O2 >>>2Al2O3
2007-04-28 20:31:58 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
10/26.98 = 0.3706449 moles
The limiting reagent is Al
4 moles Al produces 2 moles Al2O3
x/2 = 0.3706449/4
x = 0.1853225 moles Al2O3
2007-04-28 20:58:03 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
seem on the mol ratio between Al and H2 contained in the equation. are you able to observe that it really is two to three? for each 3 moles of H2 produced it only takes 2/3 of that many moles of Al So once you've 0.5 mol of H2 produced than you may want to favor 0.5 x 2/3 or 0.333 moles of Al The mass of that is: moles x molar mass 0.333 x 27 or 9 grams. all of it has to do with the mol ratio between the aspects contained in the balanced equation.
2016-12-05 01:24:45 · answer #2 · answered by ? 4 · 0⤊ 0⤋
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2016-05-17 06:18:59 · answer #3 · answered by Anonymous · 0⤊ 0⤋
4 moles of Al will react with 3 moles of O2. Atomic mass of Al = 26.98 Molecular mass of O2 = 32. 10.0g of Al = 0.371 moles, 19g of O2 = 0.594 moles Then 4*0.371 = 1.483 for Al and 3*0.594 = 1.781 The limiting reagent is Al. 4 moles of Al will produce 2 moles of Al2O3, so the moles of oxide = 0.5 times 0.371 = 0.1855 moles of Al2O3
2007-04-28 20:44:54 · answer #4 · answered by gp4rts 7 · 0⤊ 0⤋
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2007-04-28 20:34:52 · answer #5 · answered by LoverBoy 1 · 0⤊ 1⤋