A 5.00 × 10^2 g sample of Al2(SO4)3 is made to react with 4.50 × 10^2 g of Ca(OH)2. The balanced equation is: Al2(SO4)3(aq) + 3Ca(OH)2(aq) → 2Al(OH)3(s) + 3CaSO4(s) How many moles of excess reagent are unreacted?

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percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally