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%

on passing the small ampere of electric current through the tube made of ar and hg of cfl generates uv rays which excites the fluorescent and thus emits the high radiation by consumption of just 5- 12 w in comparison to incandescent light which requires more power such as around 40 - 55 w to lighten.

moreover the life span of fluorescent or cfl is around 8000 h whereas the life span of incandescent is around 1200 h

hence, the building installed with incandescent lights bears more electricity bill.