What mass of K2CO3 is required to clean up a spill of 339 ml of 2.65 M HCl if HCl(aq) reacts with K2CO3 according to

2HCl(aq) + K2CO3(aq) �¨ 2KCl(aq) + CO2(g) + H2O(l)?

To solve this problem, we need to use the balanced chemical equation and the given molarity of HCl to calculate the required mass of K2CO3.

Step 1: Write the balanced chemical equation:
2HCl(aq) + K2CO3(aq) → 2KCl(aq) + CO2(g) + H2O(l)

Step 2: Calculate the moles of HCl in the given solution:
Molarity (M) = Moles (mol) / Volume (L)

Moles of HCl = Molarity × Volume
Moles of HCl = 2.65 M × 0.339 L
Moles of HCl = 0.898 moles

Step 3: Use the balanced chemical equation to relate the moles of HCl to the moles of K2CO3.
From the balanced chemical equation, we can see that 2 moles of HCl react with 1 mole of K2CO3.

4. Calculate the moles of K2CO3 required:
Moles of K2CO3 = (Moles of HCl) / 2
Moles of K2CO3 = 0.898 moles / 2
Moles of K2CO3 = 0.449 moles

Step 5: Calculate the mass of K2CO3 required:
Molar mass of K2CO3 = (2 × atomic mass of K) + atomic mass of C + (3 × atomic mass of O)
Molar mass of K2CO3 = (2 × 39.10 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol)
Molar mass of K2CO3 = 138.21 g/mol

Mass of K2CO3 = Moles of K2CO3 × Molar mass of K2CO3
Mass of K2CO3 = 0.449 moles × 138.21 g/mol
Mass of K2CO3 ≈ 61.93 grams

Therefore, approximately 61.93 grams of K2CO3 is required to neutralize the spill.