Question

When 2.3 × 10^3 g of CaCO3 are heated, the actual yield of CaO is 1.09 × 10^3g. What is

Answer 1

The percent yield : 4. 84.58%

Further explanation

Reaction

CaCO₃ ⇄ CaO+CO₂

mass CaCO₃ = 2.3 × 10³ g

mol CaCO₃ (MW=100.0869 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{2.3\times 10^3}{100,0869}\\\\mol=22.98$

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

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

$\tt mass=mol\times MW\\\\mass=22.98\times 56,0774\\\\mass=1288.659~g$

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{1090}{1288.659}\times 100\%\\\\\5yield=84.58\%$