if 6.3 G of NaHCO3 are added to 15.0 G of CH3COOH solution the residue is found to wait 18.0 gram what is he mass of CO2 released in the reaction
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Answer:
3.3 g of CO2
Explanation:
60 g CH3COOH + 84g NaHCO3 ---> 82 g CH3COONa + 44 g CO2
Moles of NaHCO3 = 6.3/84 = 0.075
Moles of Ch3COOH = 15/60 = 0.25
NaHCO3 is the limiting reagent.
Moles of CO2 formed are = 0.075
Weight of CO2 = 3.3g
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