Question

If 6.3 g of NaHCO3 are added to 15 g of CH3COOH solution the residue Is found to weigh 18 g .what is the mass of Co 2 released in the reaction

Answer 1

Answer:

The chemical reaction will be:

NNaHCO3 +CH3 COOH→CH3COONa+H2O+CO2

COONa+H2O+CO2

molarmass:NaHCO3 =84

=84CH3COOH=60

=84CH3COOH=60CH3COONa=82

=84CH3COOH=60CH3COONa=82CCO =44

=4484g NaHCO3+60gCH3COOH→82gCH3COONa+44gCO2

COONa+44gCO2

COONa+44gCO2Moles of NaHCO= 6.3/84=0.075

=0.075Moles of CH3COOH=15/60=0.25

=0.25∴NaHCO3 is the limited reagent.

=0.25∴NaHCO3 is the limited reagent.Moles of CO2formed=0.075

=0.25∴NaHCO3 is the limited reagent.Moles of CO2formed=0.075weight of Co2 =0.075×44=3.3g

Explanation:

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