Final answer:
To neutralize 500 mL of battery acid with a concentration of 12 M, you would need approximatley 1008 grams of NaHCO₃.
Step-by-step explanation:
To determine the amount of NaHCO₃ needed to neutralize the battery acid, we need to calculate the number of moles of H₂SO₄ present in the solution. We can then use the stoichiometry of the balanced equation to find the number of moles of NaHCO₃ needed to neutralize the acid.
The balanced equation is as follows: H₂SO₄(aq) + 2NaHCO₃(aq) → Na₂SO₄(aq) + 2H₂O(l) + 2CO₂(g)
First, we need to calculate the number of moles of H₂SO₄. The concentration of the battery acid is given as 12 M. Using the formula C = n/V, where C is the concentration, n is the number of moles, and V is the volume in liters, we can rearrange the equation to solve for n.
12 M = n/(0.5 L)
n = 6 moles
According to the balanced equation, 2 moles of NaHCO₃ are required to neutralize 1 mole of H₂SO₄. Therefore, we would need 12 moles of NaHCO₃ to neutralize 6 moles of H₂SO₄. To convert moles to grams, we can use the molar mass of NaHCO₃ which is 84 g/mol.
12 moles NaHCO₃ x (84 g/mol) = 1008 g NaHCO₃
So, you would need 1008 grams of NaHCO₃ to neutralize 500 mL of the battery acid.