3.4 mol Fe(OH)3 and 7.4 mol H2SO4 react according to the equation
2 Fe(OH)3 + 3 H2SO4 −→
Fe2(SO4)3 + 6 H2O .
If the limiting reactant is Fe(OH)3, determine the amount of excess reactant that remains.

the answer is c. nacl

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