Answer to Question #317093 in General Chemistry for connor

Question

31.3 L of concentrated nitric acid (8.50 mol L^-1) was spilled on a road. Sodium carbonate solid was spread on the acid to react as in the equation:

CO₃^2-(aq) + 2H⁺ → CO₂(g) + H₂O

How many grams of sodium carbonate are needed to neutralise the acid?

Answer 1

If 1litre= 8.50mol

31.3×8.50= 266.05moles

Mole ratio= 1:2

Moles of sodium carbonate= (1/3)×266.05

= 88.68moles

Mass= 106× 88.68= 9400.43g