If 22 mL of 5.0 M H2SO4 was spilled, what is the minimum mass of NaHCO3 that must be added to the spill to neutralize the acid?
H2SO4 + 2NaHCO3 ==> Na2SO4 + 2H2O + 2CO2
mols H2SO4 = M x L = ?
mols NaHCO3 = 2xmols H2SO4
grams NaHCO3 = mols NaHCO3 x molar mass = ?
Post your work if you get stuck.
To determine the minimum mass of NaHCO3 needed to neutralize the spilled H2SO4, we need to use stoichiometry.
First, let's write a balanced chemical equation for the reaction between H2SO4 and NaHCO3:
H2SO4 + 2NaHCO3 -> Na2SO4 + 2H2O + 2CO2
From the equation, we can see that for every 1 mole of H2SO4, we need 2 moles of NaHCO3 to neutralize it.
Now, let's calculate the moles of H2SO4 in the spill:
moles of H2SO4 = volume (in L) × molar concentration
= 0.022 L × 5.0 mol/L
= 0.11 mol
Since we need a 1:2 mole ratio of H2SO4 to NaHCO3, the moles of NaHCO3 needed is twice that of H2SO4:
moles of NaHCO3 = 2 × moles of H2SO4
= 2 × 0.11 mol
= 0.22 mol
Now, we can calculate the minimum mass of NaHCO3 using its molar mass:
mass of NaHCO3 = moles × molar mass
= 0.22 mol × (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol))
≈ 15.94 g
Therefore, the minimum mass of NaHCO3 that must be added to neutralize the spilled acid is approximately 15.94 grams.