How many milliliters?

Question

How many milliliters of 3.00 M calcium hydroxide is required to neutralize 810. mL of 1.10 M hydroiodic acid?

Answer 1

Ca(OH)₂ + 2HI ==> CaI₂ + 2H₂O ... balanced equation

moles HI present = 810. ml x 1 L/1000 ml x 1.10 mol/L = 0.891 moles HI present

moles Ca(OH)₂ needed = 0.891 mol HI x 1 mol Ca(OH)₂ / 2 mol HI = 0.4455 moles Ca(OH)₂ needed

volume Ca(OH)₂ needed: (x L)(3.00 mol/L) = 0.4455 moles and x = 0.1485 L = 149 mls (3 sig. figs.)

Answer 2

Hello, Chris,

Start with balancing the equation. It tells us we need twice as many moles of HI per mole of calcium hydroxide. Next, biuld a table of what we know and what weed (the number of ml of 3 M Ca(OH)2).

Use the concentration and volume of the HI to find the moles of HI. We know that concentration of Molar means moles/1 liter. So multiple the 1.1M concentration of HI (1.1M) time the volume IN LITERS. To get liters, divide the 810ml by 1000. Once we have moles, divide that by 2 to get the number of moles of Ca(OH)2 that HI will neutralize, since that is what the balanced equation shows.

Finally, use the definition of Molar to find the number of liters that holds 0.446 moles of Ca(OH)2. 0.4455moles/X = 3. X is in liters, so multiply by 1000 to get ml.

Bob