Use the equation below to solve the problems: 4AI + 3 O_2 rightarrow 2 Al_2 O_3
ID: 507388 • Letter: U
Question
Use the equation below to solve the problems: 4AI + 3 O_2 rightarrow 2 Al_2 O_3 How main moles of AI_2 O_3 can be formed from 10.0 g of Al? How many grams of AI_2 O_3 can be formed from 10.0 g of Al? How many moles of Al_2 O_3. can be formed from 19.2 g of O_2 ? How many grams of Al_2 O_3 can be formed from 19.2 g of O_2 ? How many grams of O_2 are needed for 10.0 g of Al ? Which reactant is the limiting reactant? How many grams of Al_2 O_3 can be formed theoretically? How many grams of excess reactant remain unreacted? How many grams of Limiting reactant is left behind unreacted? Calculate the actual yield if percent yield is 66.8 %Explanation / Answer
Ans. Please note the correct for #5 and #8 in your sheet.
#5. Mass of O2 = 0.278 mol x (32.0 g mol-1) = 8.896 g
#8. Amount of O2 unreacted = 19.2 g – 8.896 g = 10.304 g
#10. Theoretical yield of Al2O3 = 18.9 g
Actual yield = 66.8 %
Now, using the formula-
% actual yield = (Experimental yield / Theoretical yield) x 100
Or, 66.8 = (Experimental yield / 18.9 g) x 100 ; [Do not write % with 66.8]
Or, 66.8 / 100 = (Experimental yield / 18.9 g)
Or, Experimental yield = 0.668 x 18.9 g = 12.6252 g
Therefore, actual/ or experimental yield of Al2O3 = 12.6252 g
#9. Actual moles of Al2O3 produced = Mass/ Molar mass
= 12.6252 g / (101.961278 g/ mol)
= 0.1238 mol
Actual moos of Al consumed = (4/ 2) x Actual moles of Al2O3 produced
= 0.2476 mol
Actual mass of Al consumed = Mass x Molar mass
= 0.2476 mol x (26.981539 g/ mol)
= 6.680 g
Mass of Al (limiting reagent) left behind = Mass of Al taken – mass of Al consumed
= 10.0 g – 6.68 g
= 3.32 g
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