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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 :

SEBASSANDINGO

Answer:

[tex]m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3[/tex]

Explanation:

Hello!

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

[tex]2NaHCO_3(s)\rightarrow Na_2CO_3(s)+H_2O(g)+CO_2(g)[/tex]

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:

[tex]m_{Na_2CO_3}^{theoretical}=10.52gNaHCO_3*\frac{1molNaHCO_3}{84.01gNaHCO_3}*\frac{1molNa_2CO_3}{2molNaHCO_3} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}\\\\ m_{Na_2CO_3}^{theoretical}=6.636gNa_2CO_3[/tex]

Best regards!

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