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How many moles of 45 g of Al03? (AI = 27) (0= 16) The molar mass of Al203 = (2 x 27) - (3 x 16) = 6 g/mol The molar mass of Al203 = (2 x 27) + (3 x 16) = 102 g/mol Mass No. of moles Molar mass 45 No. of moles = =0.44 mol 102 6. No. of moles = = 0.0588 mol 102

How many moles of 45 g of Al03? (AI = 27) (0= 16) The molar mass of Al203 = (2 x 27) - (3 x 16) = 6 g/mol The molar mass of Al203 = (2 x 27) + (3 x 16) = 102 g/mol Mass No. of moles Molar mass 45 No. of moles = =0.44 mol 102 6. No. of moles = = 0.0588 mol 102

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Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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How many moles of 45 g of Al-03? (Al = 27) (0= 16) The molar mass of Al203 = (2 x 27)- (3 x 16) = 6 g/mol The molar mass of Al203 = (2 x 27) + (3 x 16) = 102 g/mol Mass No. of moles Molar mass No. of moles = 45 = 0.44 mol 102 6. No. of moles = =0.0588 mol 102
Transcribed Image Text:How many moles of 45 g of Al-03? (Al = 27) (0= 16) The molar mass of Al203 = (2 x 27)- (3 x 16) = 6 g/mol The molar mass of Al203 = (2 x 27) + (3 x 16) = 102 g/mol Mass No. of moles Molar mass No. of moles = 45 = 0.44 mol 102 6. No. of moles = =0.0588 mol 102
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