Question

A 1.897g sample of Mg(HCO3)2 was heated and decomposed. When the sample
cooled, it weighed 1.071g. What is the % yield of this reaction?
Mg(HCO3)2 (s) + CO2 (g) + H2O (g) + MgCO2 (s)

Answer 1

1.5 % is the percent yield in the reaction.

Explanation:

Given that:

original mass of the sample used in reaction = 1.897 grams

product formed after decomposition = 1.071 grams

The reaction for the decomposition:

Mg(HCO3)2 (s) ⇒ CO2 (g) + H2O (g) + MgCO2 (s)

It says that 1 mole of Mg(HCO3)2  yielded 1 mole of  MgCO2  on decomposition

68.31 grams/mole or 68.31 grams of MgCO2 is formed

percent yield = $\frac{actual yield}{theoretical yield}$ x 100

putting the values in the equation:

percent yield = $\frac{1.071}{68.31}$

                       = 0.015 x100

   PERCENT YIELD = 1.5 %