When heated, calcium carbonate decomposes according to the equation given below: CaCO3 --> CaO + CO2
If 50.0 g of CaCO3 react but only 20.0 g of CO2 are recovered, what is the percent yield of this reaction?

90.9%
40.0%
66.7%
250%

Percent yield = 90.9%

Explanation:

Given data:

Mass of CaCO₃ = 50.0 g

Mass of CO₂ produced = 20.0 g

Percent yield = ?

Solution:

Chemical equation:

CaCO₃ →  CaO + CO₂

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 50.0 g/ 100.1 g/mol

Number of moles = 0.5 mol

Now we will compare the moles of CO₂ with CaCO₃.

                      CaCO₃          :        CO₂

                          1                :           1

                       0.5               :          0.5

Mass of CO₂: Theoretical yield

Mass = number of moles × molar mass

Mass = 0.5 mol × 44 g/mol

Mass = 22 g

Percent yield:

Percent yield = ( actual yield / theoretical yield ) × 100

Percent yield = (20.0 g/ 22.0 g) × 100

Percent yield = 0.909 × 100

Percent yield = 90.9%

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.