What volume of 0.08892 m hno3 is required to react completely with 0.2352g of potassium hydrogen phosphate? 2hno3(aq) + k2hpo4(aq)⟶h2po4(aq) + 2kno3(aq)

The volume of 0.08892 M HNO3 required to react completely with 0.2352g K2HPO4 is 0.0304 Liters

calculation

find the moles of K2HPO4 reacted

moles=mass/molar mass

=0.2352 g/174 g/mol = 1.35 x10^-3 moles

By use of mole ratio of HNO3: K2HPO4 which is 2:1 the moles of HNO3 is therefore = 1.35 x10^-3 x2 =2.7 x10^-3 moles

Volume of HNO3=moles/molarity

=2.7 x10^-3/0.08892= 0.0304 Liters

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