Final answer:
The molarity of the nitric acid that reacts with 0.198 grams of sodium carbonate is approximately 0.179 M, calculated by using the stoichiometry of the reaction and the mass of sodium carbonate.
Step-by-step explanation:
To find the molarity of the nitric acid that reacts with sodium carbonate, we must first write the balanced chemical equation:
2 HNO3 (aq) + Na2CO3 (s) → 2 NaNO3 (aq) + CO2 (g) + H2O (l)
Using this equation, we can see that it takes 2 moles of nitric acid to react with 1 mole of sodium carbonate. To proceed, we need to calculate the moles of sodium carbonate reacted:
Moles of Na2CO3 = mass (in grams) / molar mass of Na2CO3
= 0.198 g / 105.99 g/mol
≈ 0.001868 mol
Since the stoichiometry of the reaction is 2:1, the moles of HNO3 that reacted are twice that of Na2CO3:
Moles of HNO3 = 2 × moles of Na2CO3
= 2 × 0.001868 mol
≈ 0.003736 mol
Now, to find the molarity (M), we divide the moles of HNO3 by the volume of the solution in liters:
Molarity (M) = moles of solute / volume of solution (in L)
= 0.003736 mol / 0.02087 L
≈ 0.179 M
Therefore, the molarity of the nitric acid solution is approximately 0.179 M.