2NH3 + MgSO4 + 2H2O → Mg(OH)2 + (NH4)2SO4 Using the chemical equation above answer the following questions. c. If I have 4.6 moles of MgSO4 and 5.4 moles of NH3 which one is the limiting reactant? You must show your work for credit.

i think that the answer is c

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