Answer to Question #66127 in General Chemistry for Emily Gehrig

Question

What volume (in mLs) of 6.0 M HCl is required to react with 10.0 g of K2CO3?

Answer 1

1) Write down the balanced reaction equation:
2HCl + K 2 CO 3 = 2KCl + H 2 O + CO 2
From the balanced equation we can see that 1 mole of potassium carbonate requires 2 moles of HCl for complete reaction.

2) Find how many moles in 10.0 g of potassium carbonate.
Molar mass of K 2 CO 3 is (39.1 * 2) + 12.0 + (16.0 * 3) = 138.2 g/mol. So 1 mole contains 138.2 g of potassium carbonate. Than 10.0 g is 10.0 g / 138.2 g/mol = 0.07 moles. According to the balanced equation for that amount of potassium carbonate we need 0.07 mol * 2 = 0.14 moles of HCl for complete reaction.

3) Find volume of solution that contains necessary amount of acid:
Molarity of solution shows number of moles of solute per 1 liter of solution. So 1 liter of 6.0 M HCl contains 6.0 moles of HCl. Than amount of solution that contains 0.14 moles of acid is 0.14 mol / 6.0 mol/L= 0.0233 L = 23.3 mL.

Answer:
23.3 mL of HCl solution.