In the laboratory, acid spills are often neutralized by adding soldium bicarbonate. What mass of sodium bicarbonate reacts with 225 mL of 6.00 M HCl?
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To find the mass of sodium bicarbonate that reacts with 225 mL of 6.00 M HCl, we need to use the concept of stoichiometry and balanced chemical equations.
Step 1: Write the balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl):
NaHCO3 + HCl -> NaCl + H2O + CO2
Step 2: Calculate the number of moles of HCl in 225 mL of 6.00 M HCl:
Molarity (M) = moles/volume (L)
6.00 M = moles/0.225 L
moles of HCl = 6.00 M * 0.225 L = 1.35 moles
Step 3: Use the stoichiometry of the balanced chemical equation to determine the molar ratio between HCl and NaHCO3. Observation from the equation shows that 1 mole of HCl reacts with 1 mole of NaHCO3.
Step 4: Since the molar ratio between HCl and NaHCO3 is 1:1, the moles of NaHCO3 required would be the same as the moles of HCl, which is 1.35 moles.
Step 5: Calculate the molar mass of NaHCO3 (sodium bicarbonate):
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of NaHCO3 = 22.99 + 1.01 + 12.01 + (3 * 16.00) = 84.01 g/mol
Step 6: Finally, calculate the mass of sodium bicarbonate (NaHCO3) that reacts with 225 mL of 6.00 M HCl using the equation:
mass = moles * molar mass
mass = 1.35 moles * 84.01 g/mol = 113.51 g
Therefore, approximately 113.51 grams of sodium bicarbonate reacts with 225 mL of 6.00 M HCl.