Grant found the mass of an empty crucible and lid to be 62.549 g. After he added

Question

Grant found the mass of an empty crucible and lid to be 62.549 g. After he added an unknown, the mass was 63.089 g. The crucible and lid and unknown were heated strongly for 10 minutes. After cooling, the mass of the crucible and lid and sample were found to be 62.928 g. Grant also performed a flame test on a sample of his unknown, and found it produced a violet flame. Is the unknown NaHCO_3, Na_2 CO_3, KHCO_3, or K_2 CO_3? Explain the reasoning used to arrive at your answer. If 0.485 g of K_2 CO_3 were reacted according to the following equation, what would be the expected mass of the KCI product? k_2 CO_3 (s) + HCI(aq) rightarrow 2 KCI9s) + H_2 O (g) + CO_2 (g)

Explanation / Answer

Potassium Ion in a compound will color a flame violet. Further, from the molar calculation it is evident that the compound is potassium bicarbonate KHCO3.

Given, wt of unknown before heating 63.089 - 62.549 = 0.540g

and wt after heating the unknown is 62.928 - 62.549 = 0.379g

we known 2 KHCO3 = K2CO3 + CO2 + H2O

2 moles of KHCO3 gives 1 mole of K2CO3

  0.540g of KHCO3 = 0.540g/100gmol-1 = 5.4 mmol

  5.4 mmol of KHCO3 should give 2.7 mmol K2CO3 i.e., 0.0027mol x 138.205gmol-1 = 0.373 g

This is close to what has been obtained experimentally 0.006 g could be considered as an manual error. Further, similar calculation for decomposition of K2CO3 to K2O does not agree well with the observation.

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