If 13.5 ml of concentrated (17.0 M) of sulfuric acid was spilled on the chemistry counter, calculate the mass of sodium bicarbonate needed to neutralize the acid
To calculate the mass of sodium bicarbonate needed to neutralize the acid, we need to determine the amount of sulfuric acid that was spilled and then use the stoichiometry of the balanced chemical equation to find the corresponding amount of sodium bicarbonate required. Here's how you can do it:
Step 1: Determine the amount of sulfuric acid spilled in moles.
To do this, we can use the formula: moles = concentration (M) x volume (L).
Given:
Concentration of sulfuric acid (H₂SO₄): 17.0 M
Volume of sulfuric acid: 13.5 ml (milliliters) = 13.5 / 1000 L (liters) (conversion to liters is necessary)
moles of H₂SO₄ = concentration x volume
= 17.0 M x (13.5 / 1000) L
= 0.2295 moles of H₂SO₄
Step 2: Use the balanced chemical equation to find the stoichiometric ratio between sulfuric acid and sodium bicarbonate.
The balanced equation for the reaction between sulfuric acid and sodium bicarbonate is as follows:
H₂SO₄ + 2 NaHCO₃ → Na₂SO₄ + 2 H₂O + 2 CO₂
From the equation, we can see that 1 mole of sulfuric acid (H₂SO₄) reacts with 2 moles of sodium bicarbonate (NaHCO₃).
Step 3: Calculate the amount of sodium bicarbonate needed.
To do this, we can use the stoichiometric ratio obtained from the balanced chemical equation.
moles of NaHCO₃ = 2 x (moles of H₂SO₄)
= 2 x 0.2295 moles
= 0.459 moles
Step 4: Convert moles of sodium bicarbonate to mass.
To convert moles of sodium bicarbonate to mass, we need to know the molar mass of NaHCO₃, which is 84.0066 g/mol.
mass of NaHCO₃ = moles of NaHCO₃ x molar mass of NaHCO₃
= 0.459 moles x 84.0066 g/mol
= 38.509 g
Therefore, the mass of sodium bicarbonate needed to neutralize 13.5 ml of concentrated sulfuric acid is approximately 38.509 grams.