7. Below is the electrolysis of aluminum oxide. How many grams of aluminum oxide is needed...
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
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
If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start with 10.0 g of Al and 19.0 grams of O2? Reaction: 4Al+3O2 yields 2 Al2O3
How many g of Aluminum are needed to react completely with 625 gram of Mn304, according to the chemical equation below! 3 Mn304 + 8 Al -> 9 Mn + 4 Al2O3 2.65 g 19.29 24.39 1.549 1.38 g 1979 3.91 g 8.709 QUESTION 14 CS2 is reacted with oxygen according to the following equation. What mass of O2 would react completely with 9.33 grams of C527 CS2 + 302 -> CO2 + 2 502 7.91 g 173 g 4.29...
How many grams of H2O will be formed when 31.78 mol H2 is mixed with 2.64 mol of O2 and allowed to react to form water according to the reaction below? 2H2 + O2 -----> 2H2O (Note: Determine the limiting reagent first) 7.3 g 286 g 23.6 g 94.6 g 144 g
ive converted to grams of my product but not sure if i did it correctly . Can someone show the proper steps so i can compare to see my mistakes If 2.0 mol of Fe2O3(s) were mixed with 1.0 mol of Al(s) to react according to the reaction shown below, then the limiting reagent is __ _and the reagent in excess is Fe2O3(s) + 2 Al(s) — Al2O3(s) + 2 Fe(s) Fe2O3 and Al2O3 Al2O3 and Fe Ob. Ос. Al...
1. When 15.4 g of aluminum (26.98 g/mol) reacts with 14.8 g of oxygen gas (32.00 g/mol), what is the maximum mass (in grams) of aluminum oxide (101.96 g/mol) that can be produced? 4 Al(s) + 3 O2(g) → 2 Al2O3(s) 2. If 14.1 moles of Cu and 48.7 moles of HNO3 are allowed to react, how many moles of excess reactant will remain if the reaction goes to completion? Cu + 8 HNO3 → 3Cu(NO3)2 + 2NO + 4H2O...
Marked out of 1.00 Aluminum reacts with oxygen to form aluminum oxide according to the following unbalanced equation: Al(s) + O2 (g) → Al2O3 (s) If 5.0 moles of A react with excess oxygen, how many moles of Al2O3 can be produced? Select one: a. 1.0 mole b. 2.0 mole c. 3.0 mole d. 2.5 mole e. 10.0 mole Clear my choice Question 11 Not yet answered
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
1. (12 pts) For the balanced equation below, determine how many grams of UF4 (MW: 314.0 g/mol) can be produced when 7.31 kg of UO2 (MW: 270.0 g/mol) react with 3.94 L of 3.54 M HF (MW: 20.01 g/mol). UO2(g) + 4 HF(aq) → UF4(g) + 2 H20(1) 2. (12 pts) A) Determine how much chromium metal can be produced when 284.54 g of chromium (III) oxide reacts with excess aluminum according to the reaction below. Cr2O3(s) + 2 Al(n)...