When solid sodium bicarbonate (NaHCO3) is heated, it produces gaseous carbon dioxide, liquid water, and solid sodium carbonate (Na2CO3). The unbalanced equation below shows this reaction.
NaHCO3 (s) Na2CO3 (s) + CO2 (g) + H2O (l)
If 10.2 g NaHCO3 is added to a 982 mL container at 55C, what volume of water will be generated by this reaction? Recall that the density of water is 1 g/mL. Report your answer in Liters (L) using scientific notation.
The balanced equation is as follows,
2NaHCO3(s)= Na2CO3 (s) + CO2 (g) + H2O (l)
So 2 moles of NaHCO3 when consumed in heating generates one mole of H2O
The molecular weight of NaHCO3 is 84.01 g/mol whereas the molecular weight of H2O is 18 g/mol.
So in other words,
168.02 g of NaHCO3 gives 18 gm of H2O
Following, the unitary method,
10.2 g of NaHCO3 gives 1.0927 gm of H2O
The density of water is 1 g/mL
Hence the volume of water generated is 1.0927 mL= 1.0927 x 10-3 L
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