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A 78.6 g sample of CaO is reacted with 54.9 g of H2O. How many grams of the excess reactant remain after completion of the reaction? a) 0 g b) 23.7 g c) 54.9 g d) 78.6 g

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Answer:

29.7 grams, which is not one of the answer options. Perhaps the 23.7 grams was mistyped?

Step-by-step explanation:

We need a balanced equation to start. This appears to be a combination reaction:

CaO + H2O = Ca(OH)2

Let's next convert the grams of each reactant into moles. Do this by dividing the mass (g) by the molar mass for each compound (g/mole).

Reactant Grams (g) Molar Mass (g/mole) Moles

CaO 78.6 56.1 78.6g/(56.1 g/mole) = 1.40

H2O 54.9 18.0 54.9g/(18.0 g/mole) = 3.05

The balanced equation tells us that we should expect that 1 mole of H2O will be consumed for every 1 mole CaO present. We start with 1.40 moles of CaO, so that means 1.40 moles of H2O will be consumed.

Since we started with 3.05 moles of H2O, there will be unreacted H2O at the end of the reaction:

Start: 3.05 moles

Consume: 1.40 moles

Remaining: 1.65 moles

This makes H2O the excess reactant.

We want grams H2O, so convert 1.65 moles into grams. Use the molar mass of H2O:

(1.65 moles H2O)*(18.0 g/mole H2O) = 29.7 grams H2O

29.7 grams is not an answer option. It should be.

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