4 moles of Al would yield 2 moles of Al2O3
10 g Al is 10/26.98 = 0.3706 moles
You would expect 0.3706/2 moles ( 0.1853 moles) of Al2O3 at 100%
yield.
You obtained 16g/101.96 = 0.1569 moles of Al2O3
Your percent yield was 0.1569/0.1853 = 84.7%
If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the...
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?
Aluminum oxide (Al2O3) is produced according to the following equation. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) If the reaction occurs with an 82.4% yield, what mass of aluminum should be reacted with excess oxygen to produce 45.0 grams of Al2O3? a. 54.6 g Al b. 37.9 g Al c. 35.1 g Al d. 23.8 g Al e. 28.9 g Al How do we solve this problem
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 .
7. Below is the electrolysis of aluminum oxide. How many grams of aluminum oxide is needed to make 60.0 g of aluminum? Al(s) + O2(9) Al2O3(s) electrolysis with 125 g of bromine. Which is the limiting reagent according We were unable to transcribe this image
7. Below is the electrolysis of aluminum oxide. How many grams of aluminum oxide is needed to make 60.0 g of aluminum? Al2O3(s) electrolysis Al(s) + O2(g) SAL2O3 HAL+. 302 60g . 26.98 -2.22 2.22. 1. 11 mol x101.96 = 1.13 40 g of benzene is mixed with 125 g of bromine. Which is the limiting reagent according nhamical reaction below?
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
Consider the reaction for the formation of aluminum oxide from aluminum and oxygen.4Al(s)+3O2(g)⟶2Al2O3(s) ΔH11. Express the enthalpy of the following reaction, ΔH2, in terms of ΔH1.2Al2O3(s)⟶4Al(s)+3O2(g) ΔH2ΔH2=2. Express the enthalpy of the following reaction, ΔH3, in terms of ΔH1.12Al(s)+9O2(g)⟶6Al2O3(s) ΔH3 12Al(s)+9O2(g)⟶6Al2O3(s) ΔH3ΔH3=3. Express the enthalpy of the following reaction, ΔH4, in terms of ΔH1.2Al(s)+32O2(g)⟶Al2O3(s) ΔH4 2Al(s)+32O2(g)⟶Al2O3(s) ΔH4ΔH4=
For this problem, identify the limiting reagent and calculate the grams of Al2O3 obtained in the reaction of 100. grams of aluminum with 70.0 grams oxygen. If 125 grams of Al2O3 is actually produced, what is the % yield. The equations are not balanced. They were balanced in Exercise B. Use those coefficients to do these calculations. Al + O2→ Al2O3 What is the limiting reagent? ___________________ How many grams Al2O3 will be produced? ___________________ If 120 grams of Al2O3...
QUESTION 10 What is the theoretical yield of chromium (Cr) that can be produced by the reaction of 12.2 g of Cr2O3 with 3.10 g of aluminum according to the balanced chemical equation below? 2 Al + Cr2O3 - Al2O3 +2 Cr 45.8 QUESTION 11 How many grams of the excess reagent is left after the reaction has gone to completion? QUESTION 12 If the reaction produced a 65.2% yield, how many grams of chromium were produced?
How much aluminum oxide and how much carbon are needed to prepare 395 g of aluminum by the balanced chemical reaction (below) if the reaction proceeds to 84.1 % yield? 2Al2O3(s)+3C(s) 4Al(s)+3CO2(g) A_ ( ) g Al2O3 B_ ( ) g C