When aluminum metal is exsposed to oxygen gas a coating ofaluminum oxide forms o
ID: 76550 • Letter: W
Question
When aluminum metal is exsposed to oxygen gas a coating ofaluminum oxide forms on the surface of the aluminum.the balancedequation for the reaction is 4Al(s)+3O2(g) 2Al 2O3(s) Suppose a sheet of pure aluminum gaines 0.0900g of mass whenexsposed to air. Assume that this gain can be attributed to itsreaction with oxygen. a)what mass of O2 reacted with the Al? b) What mass of Al reacted? c) What mass of Al2O3 formed ? When aluminum metal is exsposed to oxygen gas a coating ofaluminum oxide forms on the surface of the aluminum.the balancedequation for the reaction is 4Al(s)+3O2(g) 2Al 2O3(s) Suppose a sheet of pure aluminum gaines 0.0900g of mass whenexsposed to air. Assume that this gain can be attributed to itsreaction with oxygen. a)what mass of O2 reacted with the Al? b) What mass of Al reacted? c) What mass of Al2O3 formed ?Explanation / Answer
The balanced equation is 4 Al(s) + 3 O2 (g) -------> 2Al2O3 (s) Mass of pure aluminm gains = 0.0900g a). As per equation 4moles of Al reacts with 3 molesof O2. So 0.0900g of Al required O2 is , = [3*(16*2) g/ 4*27g]*0.0900g = [ 3*32g/ 108]g *0.0900g = (96/108)*0.0900g = 0.08g 0.08 g of O2 is reacted with 0.0900g ofAl. b) Amount of Al reactedis 0.0900g/4*27 =7.5*10-4 g c) 4 molesof Al is formed 2.0 moles of Al2O3 So 0.0900 g of Al formed Al2O3 is = [ 2*(27*2*16*3)/4*27]*0.0900g = [ 2*( 54*48)/108]*0.0900 g = 4.32 g 4.32 g of Al2O3 isformed.Related Questions
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