a. Gram molar mass of Al2(SO4)3 = 27*2 + 32*3+16*12 = 342g/mole mass of Al2(SO4)

Question

a. Gram molar mass of Al2(SO4)3   = 27*2 + 32*3+16*12   = 342g/mole

mass of Al2(SO4)3   = no of moles of Al2(SO4)3 * gram molar mass of Al2(SO4)3

                                 = 2mole*342g/mole   = 684g >>>>>answer

b. gram molar mass of ScBr3   = 45+80*3   = 285g/mole

no of molecules     = W*6.023*10^23/gram molar mass

                                 = 4.5*6.023*10^23/285

                                  = 9.51*10^21 molecules

c. Gram molar mass of Al2(SO4)3   = 27*2 + 32*3+16*12   = 342g/mole

percentage composition of Al    = 2*atomic mass of Al *100/gram molar mass

                                                     = 2*27*100/342    = 15.79% >>>>answer

Explanation / Answer

Calculate the following (6 points). a. Number of grams in 2.0 moles of Al(SO) b. Number of molecules in 4.5 grams of ScBr c. The mass percent composition of Al in Alb(SO) al formula for an H-CI-O molecule. The percent 330 g (8 points).

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