Question

HCl reacts with Barium Hydroxide to produce barium chloride and water. how many ml of a 3.00 M hydrochloric acid solution would be required to react with 25.5 ml of a 4.65 M barium hydroxide solution?

Answer 1

To react completely with 25.5 mL of a 4.65 M barium hydroxide solution, 79.05 mL of a 3.00 M hydrochloric acid solution is required, according to the balanced chemical reaction and stoichiometry.

Step-by-step explanation:

To determine how many milliliters of a 3.00 M hydrochloric acid solution would be required to react with 25.5 ml of a 4.65 M barium hydroxide solution, we need to first write the balanced chemical equation for the reaction:

HCl (aq) + Ba(OH)2 (aq) -> BaCl2 (aq) + 2H2O (l)

From the equation, we see that two moles of HCl are required to react with one mole of Ba(OH)2. So, we need to calculate the number of moles of Ba(OH)2 we have:

Number of moles of Ba(OH)2 = Volume (L) × Molarity (M)
= 0.0255 L × 4.65 M
= 0.118575 mol

Since the stoichiometry of the reaction requires two moles of HCl per mole of Ba(OH)2, we need

Number of moles of HCl needed = 0.118575 mol × 2
= 0.23715 mol

Finally, we use the molarity of the HCl solution to find out the volume needed:
Volume of HCl solution (L) = Number of moles of HCl ÷ Molarity of HCl
Volume of HCl solution (L) = 0.23715 mol ÷ 3.00 M
= 0.07905 L

Convert liters to milliliters:

Volume of HCl solution (mL) = 0.07905 L × 1000 mL/L
= 79.05 mL

Answer 2

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