When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is the percent yield?
1. 51.0643
2. 54.084
3. 53.0474
4. 84.5827
5. 59.4924
6. 93.7005
7. 49.4244
8. 78.3748
9. 44.6193
10. 72.7638
Answer in units of %.

The percent yield : 4. 84.58%

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

From the equation, mol CaCO₃ : CaO = 1 : 1, so mol CaO=22.98

mass CaO(MW=56.0774 g/mol)⇒ (theoretical) :

The percent yield :

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