Given the following reaction: 2 C3H18 + 25 O2 → 16 CO2 + 18 H20 What is the percent yield when 41.43 g of CO2 are formed from the
reaction of 2.09 mole of C3H18 reacts with 2.22 mole of O2?

it is a solid

explanation:

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