O A. -8377.5 kJ Consider the following reaction: 4Al (s) + 302 (g) → 2 Al2O3...
Consider the following reaction at 298 K. 4Al(s)+3O2(g)⟶2Al2O3(s) Δ?∘=−3351.4 kJ Calculate the following quantities. Refer to the standard entropy values as needed. Δ?sys= J/K Δ?surr= J/K Δ?univ= J/K Al Entropy = 28.3 O2 Entropy=205.2 Al2O3= 50.9
Question 3 10 points The reaction 4Al (s)+302 (g) 2 Al203 (s) AH 851 k is and therefore heat is by the reaction. exothermic, absorbed endothermic, absorbed endothermic, released exothermic, released thermoneutral, neither released nor absorbed
4. Consider the reaction of aluminum and oxygen: Al(s) + O2(g) Al2O3(s) (i). Which is the limiting reactant if we start with 30.0 g Al and 30.0 g O2? (ii). What is the Theoretical Yield for the reaction? (iii). If 25.85 g of Al2O3 was collected at the completion of the reaction (actual yield), what is the % yield for the reaction? 5. (a). A 1.506-g sample of limestone-containing material gave 0.558 g of...
Given 2Al(s) + (3/2)O2(g) → Al2O3(s), AH°f = –1,670 kJ/mol for Al2O3 (s). Determine AH° for the reaction 3Al2O3(s) →6Al(s) + (9/2)02(g). Write answer to three significant figures. Numeric Response
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
The overall reaction in a commercial heat pack can be represented as 4Fe(s) + 302(g). 2Fe,03(s) -1652 kJ a. How much heat is released when 4.80 moles of iron are reacted with excess O2? Heat kJ b. How much heat is released when 1.00 mole of Fe2Os is produced? Heat = kJ c. How much heat is released when 1.70 g iron is reacted with excess O2? Heat kJ d. How much heat is released when 10.6 g Fe and...
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?
Part A Several reactions and their standard reaction enthalpies at 298.15 K are given here: Al_C3(s) + 12H2O(l) + 4Al(OH),(s) + 3CH4(g) 2Al(s) + O2(g) + Al2O3(s) 1 A1,03(s) + H2O(l) + Al(OH)3(s) AH (kJ. mol-?) –1683.0 -1675.7 -9.6 The standard enthalpies of combustion of graphite and CH4(8) are -393.51 and --890.35 kJ. molº respectively. Calculate the standard enthalpy of formation of Al4C3(s) at 25°C.
Consider the two reactions. 2 NH3(g) + 3N2O(g) + 4N2(g) + 3H2O(1) 4 NH3(g) + 302(g) + 2N2(g) + 6H2O(1) AH° = -1010 kJ AH° = 1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N,(8) + O2(8) + N20(8) AH° = k.
Consider the reaction 2H2S(g) 302(g) >2H20(I) + 2SO2(g) -1.124x103 kJ and AS° = -390.7 J/K at 298.15 K for which AHo (1) Calculate the entropy change of the UNIVERSE when 2.258 moles of H2s(g) react under standard conditions at 298.15 K ASuniverse J/K (2) Is this reaction reactant or product favored under standard conditions? 3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored...