Question

. How many grams of sodium bicarbonate are
required to neutralize 10mL of 0.902M vinegar.

Answer 1

Explanation:

CH3COOH+NaHCO3→CH3COONa+H2O+CO2

Equivalents of CH3COOH=Molarity×volume×nf=0.902×10×10−3

Equivalents of NaHCO3 required =Moles×nf=0.902×10×10−3

Moles of NaHCO3=9.02×10−3

Weight of NaHCO3=9.02×10−3×84=0.75 grams

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