What volume of .0500 M Ba(OH)2 will react completely with 21.00 mL of .500 M HCl?
Same type problem. Post your work if you get stuck.
2HCl + Ba(OH)2 --> 2HOH + BaCl2
(.0500 mol HCl / 1000 mL) (21.00 mL) (1 mol Ba(OH)2 / 2 mol HCl) (1000 mL / .0500 mol Ba(OH)2)
= 1.05e2 mL Ba(OH)2
right.
To calculate the volume of 0.0500 M Ba(OH)2 required to react completely with 21.00 mL of 0.500 M HCl, we can use the stoichiometry of the balanced chemical equation:
2HCl + Ba(OH)2 → 2H2O + BaCl2
The ratio of moles of HCl to Ba(OH)2 is 2:1, which means that 2 moles of HCl react with 1 mole of Ba(OH)2.
First, calculate the moles of HCl in 21.00 mL of 0.500 M HCl:
Moles of HCl = concentration (M) × volume (L)
= (0.500 mol/L) × (21.00 mL/1000 mL)
= 0.0105 mol
Now, use the stoichiometry to find the moles of Ba(OH)2 that would react completely with the moles of HCl:
Moles of Ba(OH)2 = (2/1) × 0.0105 mol
= 0.0210 mol
Finally, convert the moles of Ba(OH)2 to volume using its concentration:
Volume of Ba(OH)2 = Moles of Ba(OH)2 / Concentration of Ba(OH)2
= 0.0210 mol / 0.0500 mol/L
= 0.420 L
Since the given concentration was in units of Molarity (mol/L), we need to convert the final volume to milliliters:
Volume of Ba(OH)2 = 0.420 L × 1000 mL/L
= 420 mL
Therefore, the volume of 0.0500 M Ba(OH)2 required to react completely with 21.00 mL of 0.500 M HCl is 420 mL.