Question

When NaHCO3 completely decomposes, it can follow this balanced chemical equation:

Answer 1

The theoretical yield of $Na_2CO_3$ will be 2.0458 grams while that of $H_2CO_3$ will be 1.1966 grams

Stoichiometric calculation

From the equation of the reaction, the mole ratio of $NaHCO_3$ to $Na_2CO_3$  to   $H_2CO_3$ is 2:1:1.

Mole of 3.24 grams  $NaHCO_3$ = 3.24/84 = 0.0386 moles

Equivalent mole of  $Na_2CO_3$ and $H_2CO_3$ = 0.0386/2 = 0.0193 moles

Mass of 0.0193 moles $H_2CO_3$ = 0.0193 x 62 = 1.1966 grams

Mass of 0.0193 moles  $Na_2CO_3$ = 0.0193 x 106 = 2.0458 grams

The only solid product in the decomposition of $NaHCO_3$  is  $Na_2CO_3$ . Thus, it has to be the product with 2.01 grams as mass. Other products in the decomposition are gases.

Percent yield of  $Na_2CO_3$  = 2.01/2.0458 = 98%

More on stoichiometric calculations can be found here: brainly.com/question/8062886