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MOLES = MASS/GFM Al

PROBLEM 4 OF LIMITING REAGENT SET. 4 Al + 3 O 2  2 Al 2 O 3 FIRST DETERMINE IF THERE IS A LIMITING REAGENT . FIND MOLES OF ALUMINUM METAL FIND MOLES OF OXYGEN GAS. MOLES = MASS/GFM Al. MOLES = 2.5g Al/ 26.98g/mol. MOLES = 0.092661mol Al. MOLES = MASS/GFM O 2.

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MOLES = MASS/GFM Al

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  1. PROBLEM 4 OF LIMITING REAGENT SET.4 Al + 3O2 2Al2O3 FIRST DETERMINE IF THERE IS A LIMITING REAGENT.FIND MOLES OF ALUMINUM METAL FIND MOLES OF OXYGEN GAS. MOLES = MASS/GFM Al MOLES = 2.5g Al/ 26.98g/mol MOLES = 0.092661mol Al MOLES = MASS/GFM O2 MOLES = 2.5g O2 / 32.00g/mol MOLES = 0.0781mol O2 RATIO BOTH REACTANTS TO FIND LIMITING REAGENT. O23 = 0.07810 mol O2 Al 4 0.09266 mol Al 0.75 = .842 The actual ratio is larger than the theoretical, therefore the actual ratio fraction is too large, the denominator is to small Al is limiting.

  2. 4 Al + 3O2 2Al2O3 STEP 1 Identify known (the substance that can be converted to moles), choose the correct equation to convert to moles. MOLES = MASS/GFM Al MOLES = 2.5g Al/ 26.98g/mol MOLES = 0.092661mol Al STEP 2 RATIO known to objective, Al=4=0.092661 , Al2O32XX = 0.0463. mol Al2O3 STEP 3 Convert Objective to required units. % yield = experimental/theoretical 3.5 / 4.7236 =74% MOLES = MASS/GFM Al2O3 .0463 mol= mass/ 101.96 Mass =4.7236 g Al2O3

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