What mass of K2CO3 is required to clean up a spill of 282 ml of 1.05 M HCl if HCl(aq) reacts with K2CO3 according to
2HCl(aq) + K2CO3(aq) → 2KCl(aq) + CO2(g) + H2O(l)?
To find the mass of K2CO3 required to neutralize 1.05 M HCl and clean up the spill, we need to follow these steps:
Step 1: Write down the balanced chemical equation:
2HCl(aq) + K2CO3(aq) → 2KCl(aq) + CO2(g) + H2O(l)
Step 2: Determine the molar ratio between HCl and K2CO3 from the balanced equation:
According to the equation, 2 moles of HCl react with 1 mole of K2CO3.
Step 3: Calculate the moles of HCl:
To find the moles of HCl, we need to use the formula: moles = concentration (mol/L) × volume (L).
Given that the concentration of HCl is 1.05 M and the volume is 282 ml (or 0.282 L), we can calculate the moles of HCl as follows:
moles of HCl = 1.05 M × 0.282 L = 0.2951 mol
Step 4: Use the molar ratio to find the moles of K2CO3:
From the balanced equation, we know that 2 moles of HCl react with 1 mole of K2CO3. Therefore, we can determine the moles of K2CO3 as follows:
moles of K2CO3 = (moles of HCl) / 2 = 0.2951 mol / 2 = 0.14755 mol
Step 5: Convert the moles of K2CO3 to grams:
To convert moles to grams, we need to use the molar mass of K2CO3. The molar mass of K2CO3 = 2(39.10 g/mol for K) + 12.01 g/mol for C + 3(16.00 g/mol for O).
molar mass of K2CO3 = 2(39.10) + 12.01 + 3(16.00) = 138.21 g/mol.
Finally, we can calculate the mass of K2CO3 required:
mass of K2CO3 = (moles of K2CO3) × (molar mass of K2CO3)
mass of K2CO3 = 0.14755 mol × 138.21 g/mol = 20.36 g.
Therefore, approximately 20.36 grams of K2CO3 is required to clean up the spill of 282 ml of 1.05 M HCl.