Balanced chemical equation is:
4 Al(s) + 3 O2 (g) → 2 Al2O3 (s)
From above reaction,
mol of Al2O3 produced = (2/4)*number of mol of Al
= (2/4)*5.0 mol
= 2.5 mol
Answer: d
Marked out of 1.00 Aluminum reacts with oxygen to form aluminum oxide according to the following...
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)
1. Aluminum reacts with oxygen according to the following reaction: 4 Al(s) + 3 O2(g) -> 2 Al2O3 If 8.00 moles of oxygen are reacted with 8.00 moles of aluminum to completion, which is the excess reactant?
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...
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...
Question 4 Not changed since last attempt Marked out of 1.00 Calculate w (in kJ) when 189 g iron (III) oxide (MM = 159.7 g/mol) reacts with excess carbon to produce carbon dioxide gas at 451 K: 2Fe2O3(s) + 3C(s) + 4Fe(s) + 3CO2(g) P Flag question Answer: Question 5 Not yet answered Calculate the work (in kJ) when 2.20 moles of methane react with excess oxygen at 470 K: Marked out of 1.00 CH4(g) + 20 (g) + CO2(g)...
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
Iron reacts with oxygen to form iron(III) oxide according to the chemical equation below. What is the theoretical yield of product when 5.00 grams of Fe react with excess O2? [Molar masses: Fe, 55.85 g/mol; O, 16.00g/mol] 5 pts 4Fe(s) + 3O2(g) → 2Fe2O3(s)
Question 5 Solid aluminum and gaseous oxygen react in a combination reaction to produce aluminum oxide: 4Al(s) + 302 (g) → 2A1203 (s) The maximum amount of A1203 that can be produced from 2.5 g of Al and 2.5 g of O2 is 5.0 ③ 4.7 3 9.4 * 5.3 * 7.4 Question 3 -/1 The combustion of ammonia in the presence of excess oxygen yields NO2 and H20: 4 NH3(g) + 7 02 (g) → 4NO2(g) + 6H20 (g)...
Iron oxide reacts with aluminum in an exothermic reaction. Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s) The reaction of 5.00g Fe2O3 with excess Al(s) evolves 26.6 kJ of energy in the form of heat. Calculate the enthalpy change per mole of Fe2O3 reacted.