Question

What is the percent yield of the following reaction if 20 grams of CaCO3 is heated to give 5 grams of CaO? CaCO3→CaO + CO2

Answer 1

Answer:

The percentage yield of the reaction is 56.82 %.

Explanation:

The balanced chemical equation for the reaction can be written as:

$CaCO_{3}$ → $CaO + CO_{2}$

Given that :

mass of $CaCO_{3}$ = 20 g

mass of $CO_{2}$ = 5 g

Molar mass of $CaCO_{3}$ = (1×40) + (1×12) + (3×16)

                                    = 100 g/mol

Number of moles of $CaCO_{3}$ = Given mass ÷ molar mass

                                              = $\frac{20}{100}$

                                              = 0.2 moles

According to the above chemical equation,

1 mole of $CaCO_{3}$ gives 1 mole of $CO_{2}$

∴0.2 moles of $CaCO_{3}$ gives 0.2 moles of $CO_{2}$

⇒theoritical mass of $CO_{2}$ = 0.2 × (molar mass of $CO_{2}$)

                                          = 0.2 × 44

                                          = 8.8 g

Yield percentage = $\frac{actaul weight}{theoritical weight}$ × 100

                             = $\frac{5}{8.8}$× 100

                             = 56.82 %