Question

Ca + 2 H₂O ------> Ca(OH)₂ + H₂

If we have 4 moles of Ca and 4 moles of water – how much Calcium Hydroxide can we make? What would be the limiting reagent and what would be the one in excess?

Answer 1

First, figure out which the limiting reagent is. You do this by finding out how many moles of water are in it's 7.77grams, and how many moles of CaO there are in it's 4.44grams. Water is 18grams/mol. One mole of hydrogen is ~1gram, and one mole of oxygen is ~16 grams. Thus two moles of H and one mole of O give 18grams. So, 7.77grams / 18 grams/mol = 0.43 moles of water.

Next, do the same for calcium oxide. This weighs 56 grams/mole. So 4.44grams / 56grams/mole = 0.08 moles. Now you have both your reagents on an even playing field, you compare the number of moles. Since you have 0.43 moles of water, compared to 0.08 moles of calcium oxide, you will run out of the calcium oxide before you run out of water. Therefore, the calcium oxide is the limiting reagent. That makes water the excess reagent.

Once you know the limiting reagent, you use THAT value to determine how much calcium hydroxide you will end up with. Remember, the point of this excercise is to show you that you need BOTH CaO and H2O to react to form CaOH2. Once you run out of one of the reagents, the rxn will no longer ensue. That said, if you were to calculate the amount of calcium hydroxide using the moles of water, it would be innaccurate because so much of the water is in excess and goes unused. This is an important concept in chemistry. So, we'll use the CaO.

IT's a 1:1 mole ratio...which means one mole of CaO plus one mole of H20 gives one mole of CaOH2. It's very straightforward. CaOH2 weighs 74grams per mole. So we get -

4.44grams is ALL of the CaO we have, which = 0.08moles of CaO.

1 mole of CaO will yield 1 mole of CaOH2 (Look at the balanced equation you posted!). Therefore, since we only have 0.08moles of CaO, we will only get 0.08 moles of CaOH2.

All you need to do now is convert 0.08 moles of CaOH2 into grams.
0.08 moles * (74grams / mole) = 5.92 grams.

And this answer makes sense, since CaO is comparable in molar mass to CaOH2. CaO weighs 75% of what CaOH2 weighs. Thus, CaO is also about 75% of the total CaOH2 formed at the rxn's completion. (4.44g / 5.92g ) = .75, (56g/mol / 74g/mol) = .75