What is the molarity of a nitric acid solution if 25.00 mL of HNO₃ is required to neutralize 0.500 g of calcium carbonate (100.09 g/mol)? 2 HNO₃(aq)+ CaCO₃(s)→ Ca(NO₃)2(aq)+ H₂O(l)+ CO₂(g)

Question

Quizplus · Accepted Answer

First, calculate the number of moles of calcium carbonate:
0.500 g CaCO₃ x (1 mol CaCO₃ / 100.09 g CaCO₃) = 0.004998 mol CaCO₃
From the balanced equation, we know that 2 moles of nitric acid are required to neutralize 1 mole of calcium carbonate. Therefore, we need 2 x 0.004998 = 0.009996 mol HNO₃ to neutralize the CaCO₃.
Finally, calculate the molarity of the HNO₃ solution:
Molarity = moles of solute / volume of solution in liters
Assuming the HNO₃ solution is the limiting reagent and therefore we used 25.00 mL = 0.02500 L of it:
Molarity = 0.009996 mol / 0.02500 L = 0.3998 M ≈ 0.400 M.

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