When exposed to air, aluminum metal, Al, reacts with oxygen, O2, to produce a protective coating of aluminum oxide, Al2O3, which prevents the aluminum from rusting underneath. The balanced reaction is shown here:
4Al+3O2→2Al2O3
a. What is the theoretical yield of aluminum oxide if 1.60 mol of aluminum metal is exposed to 1.50 mol of oxygen?
Balanced chemical equation is:
4 Al + 3 O2 ---> 2 Al2O3
4 mol of Al reacts with 3 mol of O2
for 1.6 mol of Al, 1.2 mol of O2 is required
But we have 1.5 mol of O2
so, Al is limiting reagent
we will use Al in further calculation
Molar mass of Al2O3,
MM = 2*MM(Al) + 3*MM(O)
= 2*26.98 + 3*16.0
= 101.96 g/mol
According to balanced equation
mol of Al2O3 formed = (2/4)* moles of Al
= (2/4)*1.6
= 0.8 mol
use:
mass of Al2O3 = number of mol * molar mass
= 0.8*1.02*10^2
= 81.57 g
Answer: 81.6 g
When exposed to air, aluminum metal, Al, reacts with oxygen, O2, to produce a protective coating of aluminum oxide, Al2O...
When exposed to air, aluminum metal, Al, reacts with oxygen, O2, to produce a protective coating of aluminum oxide, Al2O3, which prevents the aluminum from rusting underneath. The balanced reaction is shown here: 4Al+3O2→2Al2O3 In Part A, we saw that the theoretical yield of aluminum oxide is 1.90 mol . Calculate the percent yield if the actual yield of aluminum oxide is 1.18 mol .
Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion, 4Al(s) + 3O2(g) → 2Al2O3(s). ΔG° for this reaction is ___________ and this reaction is ____________ . ΔG°f (kJ/mol) Al(s) 0 O2(g) 0 Al2O3(s) -1576.4 Select one: a. 0 kJ/mol; at equilibrium b. 3152.8 kJ/mol; spontaneous c. 3152.8 kJ/mol; nonspontaneous d. -3152.8 kJ/mol; spontaneous e. -3152.8 kJ/mol, nonspontaneous
X 1. Aluminum reacts with oxygen to produce aluminum oxide 4 Al(s) + 3 O2(g) + 2 Al2O3(s) IF 3.0 moles of Alreact with excess O2, how many moles of AlzO3 can be formed? a) 1.0 mol b) 1.50 mol c) 3.0 mol d ) 4.0 mol e) 4.5 mol 10. What is the log of 5,400.05? a) 3.07 b) 3.37 c) 3.73 d) 37.32 e) 37.23 Short Answer Questions (True or False)
Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s) What volume of O2 gas (in L), measured at 775 Torr and 23 ∘C , completely reacts with 52.1 g of Al?
Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s) What volume of O2 gas (in L), measured at 782 Torr and 29 ∘C , completely reacts with 52.6 g of Al? (in L)
Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s) What volume of O2 gas (in L), measured at 774 Torr and 29 ∘C , completely reacts with 52.0 g of Al? (answer in Litres)
Oxygen gas reacts with powdered aluminum according to the following reaction: 4Al(s)+3O2(g)?2Al2O3(s). What volume of O2 gas, measured at 793 mmHg and 28 ?C, is required to completely react with 53.5 g of Al? Express the volume in liters to three significant figures.
Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s) Part A What volume of O2 gas (in L), measured at 762 Torr and 32 ∘C , completely reacts with 52.1 g of Al?
Oxygen gas reacts with powdered aluminum according to the reaction: 4Al(s)+3O2(g)→2Al2O3(s) Part A What volume of O2 gas (in L), measured at 771 mmHg and 34 ∘C, is required to completely react with 51.9 g of Al?
help me do the correct math 25. (12 pts) Aluminum oxide forms when aluminum-reacts with oxygen. Balance the following chemical equation (the molar mass of wluminum oxide is 101.96 g/mol): 4 Al+ 3 02 2 Al2O3 62.79 g of aluminum metal is allowed to react. How many moles of aluminum is this? 52.79 gray.101.96 glmolo = 6402 moles Al How many moles of Al2O3 can be produced from the previously calculated moles of aluminum? 23 6402 x 6o2to² 3 =31858...