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Aluminum reacts with oxygen gas to produce a single solid compound. If 16g of aluminum react with 24g of oxyg?

Aluminum reacts with oxygen gas to produce a single solid compound. If 16g of aluminum react with 24g of oxygen, a)write the balanced equation.b)Determine the limiting reactant and the excessc)How many grams can be produced?d)How many molecules of the product can be produced?e)How many moles of the reactant in excess actually react?f)How many moles of the reactant in excess do NOT react? ( how much is in excess?)g)How many grams are present of the reactant in excess?h)If 8.1 grams of the product are actually produced, what is your percent yield?

Answer:

you opt for to understand the proscribing reactant= the reactant that produces smallest quantity of product the smallest quantity of product is the finest product 219 g AlF3/a million * a million mol AlF3/ 80 3.977 g AlF3 * 6 mol F2/ 4 mol AlF3 * 37.997 g F2/ a million mol F2= 149 g F2 40 8 g O2/ a million * a million mol O2/ 31.999 g O2 * 6 mol F2/ 3 mol O2* 37.997 g F2 / a million mol F2= 114 g F2 oxygen gasoline is the limitng reactant aluminum flouride is the added reactant some is ate up different there to verify all proscribing reactant is ate up
a) 4Al(s) + 3O2(g) ---------- 2Al2O3(s) b) Aluminum has an atomic mass of 27 g/mole, whilst oxygen has an molecular mass of 32g/mole. From Equation in a 4 moles of aluminum will react with three moles of oxygen Therefore 16g possessing 16/27= 0.5926moles will react with 3*0.5926/4= 0.4444 moles of oxygen. This will convert to 0.4444*32g of oxygen= 14.222g of oxygen. Therefore aluminum is the limited reactant and oxygen is the excess. c) The molecular mass of the aluminum oxide is (27*2+3*16)=102g/mole. 0.5926 moles of aluminum will produce 0.5926/2 = 0.2963 moles of aluminum oxide. This converts to 0.2963*102= 30.2226g of aluminum oxide. d) Two moles of the product is produced. e). 0.4444 moles of Oxygen react (see above) f) Mass of oxygen in excess is 24-14.222= 9.7778 g. Dividing this by the molecular mass of oxygen = 9.7778/32=0.3056 moles. g) 9.7778g of oxygen. h) yield = actually obtained/ theoretically obtained * 100% =8.1/30.2226 *100 = 26.80%

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