Question

1. For some reactions having trace amounts of water in your solvent can be a problem. In drier states like Arizona you can get away with buying dry solvents. However, in Florida as soon as you open the bottle up the solvent is saturated with water due to the humidity. Therefore to do water sensitive chemistry in Florida you need to dry the solvents yourself just before use. Often you’ll do this using a commercially available Solvent Purification System. However, if you need to use a solvent that you do not already have as part of your Solvent Purification System, you’ll need to distill it with a drying agent. In my lab I have had to do this for heptane. The best way to dry heptane is to add calcium hydride to a flask full of heptane and distill off the dry heptane.

   1. The water in the heptane will react with calcium hydride in a single replacement reaction where one of the products is hydrogen gas. Write the balanced chemical equation for the reaction of calcium hydride and water.

   2. When I’ve used this reaction to dry a quantity of heptane in my lab, I use an excess of calcium hydride. (Around 20.0 g usually) If I were to dump in this 20.0 g of calcium hydride into a 700 mL sample of heptane and have collected 1.23 g of hydrogen gas, what would be the concentration (in M) of water in the original 700 mL sample of heptane before drying?

   3. Based on the amount of hydrogen gas collected in part b, what mass of calcium hydride would remain unreacted after the heptane sample is dry?

Answer 1

Part-A:

Balanced chemical reaction equation is: CaH₂ + 2 H₂O → Ca(OH)₂ + 2 H₂

molar mass of CaH₂ = 42.09388 g/mol

molar mass of H₂ = 2.01588 g/mol

moles present in 20.0 g of CaH₂ = 20.0 g / 42.09388 g/mol = 0.47512845097 mol

moles present in 1.23 g of H₂ = 1.23 g / 2.01588 g/mol = 0.61015536639 mol

1 mol CaH2 and 2 mol H₂O reacts together and forms 2 mol H₂ gas

moles water H₂O = moles H₂ = 0.61015536639 mol

solution volume = 700 mL = 0.7 L

concentration (in M) of water in the original 700 mL sample of heptane before drying = 0.61015536639 mol / 0.7 L = 0.87165052341 M

Answer: concentration (in M) of water in the original 700 mL sample of heptane before drying = 0.872 M

-------------------------------------------------------------------------------------------------------------------

Part-B:

1 mol CaH₂ = 2 mol H₂ gas

moles CaH2 = moles H₂ / 2 = 0.61015536639 mol / 2 = 0.30507768319 mol CaH₂ consumed

moles of calcium hydride unreacted after the heptane sample is dry = 0.47512845097 mol - 0.30507768319 mol = 0.17005076778 mol

mass of calcium hydride would remain unreacted after the heptane sample is dry = 0.17005076778 mol x 42.09388 g/mol = 7.15809661284 g

Answer: mass of calcium hydride would remain unreacted after the heptane sample is dry = 7.16 g

-------------------------------------------------------------------------------------------------------------------

Hope this helped you!

Thank You So Much! Please Rate this answer as you wish.("Thumbs Up")