How many grams of baking soda would be needed to neutralize one litter of battery acid that has been spilled on your courage floor assume that a concentrate of the battery acid is 10 molar.?

To neutralize one liter of 10 molar battery acid, we need to calculate the amount of baking soda required to react with all the acid present.

The molar mass of baking soda (NaHCO3) is approximately 84 grams/mol.

First, we need to determine the number of moles of acid present in one liter of 10 molar battery acid:

10 moles/L * 1 L = 10 moles

Since the reaction between baking soda and battery acid is 1:1, we need an equal number of moles of baking soda to neutralize the acid. Therefore, we need 10 moles of baking soda.

Now, we can calculate the mass of baking soda needed:

10 moles * 84 grams/mol = 840 grams

Therefore, 840 grams of baking soda would be needed to neutralize one liter of 10 molar battery acid spilled on your garage floor.

Step 1: Determine the number of moles of battery acid.

Molarity (M) = moles of solute / volume of solution (in liters)
M = 10 mol/L
V = 1 L

Moles of battery acid = M x V = 10 mol/L x 1 L = 10 mol

Step 2: Determine the neutralization reaction.

Battery acid is typically sulfuric acid (H2SO4). The neutralization reaction with baking soda (NaHCO3) is as follows:

H2SO4 + 2NaHCO3 → Na2SO4 + 2H2O + 2CO2

Step 3: Calculate the grams of baking soda needed.

From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3. Therefore, 10 moles of H2SO4 will react with 20 moles of NaHCO3.

Molar mass of NaHCO3 = 84.01 g/mol

Grams of NaHCO3 = moles x molar mass = 20 mol x 84.01 g/mol = 1680.2 g

Therefore, approximately 1680.2 grams of baking soda would be needed to neutralize one liter of 10 molar battery acid.