Question

A student heats 10.52 g of sodium hydrogen carbonate in a crucible until the compound completely decomposes to sodium carbonate (rxn 1 in your lab write-up). Calculate the theoretical yield (in grams) of sodium carbonate for this reaction.

Answer 1

`m_(Na_2CO_3)^(theoretical)=6.636gNa_2CO_3`

Step-by-step explanation:

Hello!

In this case, since the decomposition of sodium hydrogen carbonate is:

`2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)`

Thus, since there is a 2:1 mole ratio between the sodium hydrogen carbonate and sodium carbonate, and the molar masses are 84.01 and 105.99 g/mol respectively, we obtain the following theoretical yield:

`m_(Na_2CO_3)^(theoretical)=10.52gNaHCO_3*(1molNaHCO_3)/(84.01gNaHCO_3)*(1molNa_2CO_3)/(2molNaHCO_3) *(105.99gNa_2CO_3)/(1molNa_2CO_3)\\\\ m_(Na_2CO_3)^(theoretical)=6.636gNa_2CO_3`

Best regards!