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Sever21 [200]
2 years ago
9

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

Chemistry
1 answer:
Hoochie [10]2 years ago
7 0

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

<h3>Stoichiometric calculation</h3>

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

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