Sodium bicarbonate is sometimes used to neutralize acid spills. If 2.0 L of 9.0 M H2SO4 is spilled, what mass of sodium bicarbonate (NaHCO3) is required to neutralize all the H2SO4?
To determine the mass of sodium bicarbonate required to neutralize the H2SO4, we need to calculate the amount of moles of H2SO4 spilled and then use the stoichiometry of the reaction to find the amount of NaHCO3 needed.
Here's how you can calculate it step-by-step:
Step 1: Calculate the number of moles of H2SO4.
The given volume of H2SO4 is 2.0 L, and the concentration is 9.0 M. To find the number of moles, we can use the formula:
moles = volume (in liters) x concentration (in moles per liter)
moles of H2SO4 = 2.0 L x 9.0 mol/L
moles of H2SO4 = 18.0 mol
Step 2: Write the balanced chemical equation for the neutralization reaction.
The balanced equation between H2SO4 and NaHCO3 is:
H2SO4 + 2NaHCO3 -> Na2SO4 + 2H2O + 2CO2
According to the balanced equation, it takes 1 mole of H2SO4 to react with 2 moles of NaHCO3.
Step 3: Calculate the number of moles of NaHCO3 required.
Since the stoichiometry says that 1 mole of H2SO4 reacts with 2 moles of NaHCO3, we can use the ratio to calculate the required amount of NaHCO3.
moles of NaHCO3 = (moles of H2SO4) x (2 moles of NaHCO3 / 1 mole of H2SO4)
moles of NaHCO3 = 18.0 mol x 2/1
moles of NaHCO3 = 36.0 mol
Step 4: Calculate the mass of NaHCO3.
The molar mass of NaHCO3 is around 84.0 g/mol.
mass of NaHCO3 = (moles of NaHCO3) x (molar mass of NaHCO3)
mass of NaHCO3 = 36.0 mol x 84.0 g/mol
mass of NaHCO3 = 3024 g
Therefore, approximately 3024 grams of sodium bicarbonate (NaHCO3) is required to neutralize all the H2SO4 in a 2.0 L spill of 9.0 M H2SO4.
2NaHCO3 + H2SO4 ==> Na2SO4 + 2H2O + 2CO2
mols H2SO4 spilled = M x L = 9.0 x 2.0 = 18
From the equation, 2 mols NaHCO3 are needed for each mol H2SO4 so you will need 36 mols NaHCO3.
M NaHCO3 = mols/L. You know M and mols, solve for L.
Post your work if you get stuck.