Question

4.8g of calcium is added to 3.6g of water. The following reaction occurs

Ca + 2H2O  Ca(OH)2 + H2
i. the number of moles of calcium = __________________
ii. the number of moles of water = ____________________
iii. Which reagent is in excess? Explain your choice.

iv. Calculate the mass of the reagent named in (ii) which remained at the end of the
experiment.

Answer 1

Reaction of interest is: Ca + 2H2O ↔ Ca(OH)2 + H2

Answer 1:
Number of moles of calcium =
`\frac{\text{weight of Ca (g)}}{\text{Atomic Weight}}`

Atomic weight of Ca = 40.078 g/mol

Therefore, Number of moles of calcium =
`\frac{\text{4.8}}{\text{40.078}}`
= 0.1198.

Number of moles of Ca = 0.1198.
....................................................................................................................

Answer 2:
Number of moles of water =
`\frac{\text{weight of H2O (g)}}{\text{Molecular Weight}}`

Molecular weight of H2O = 18 g/mol

Therefore, Number of moles of water =
`\frac{\text{3.6}}{\text{18}}`
= 0.2.

Number of moles of water = 0.2.
......................................................................................................................

Answer 3:
Number of moles of water = 0.2
Number of moles of Ca = 0.1198

Here it can be seen that, Number of moles of water > Number of moles of Ca.
Hence, water is in excess.
.........................................................................................................................

Answer 4:
Number of moles of water that is in excess = 0.2 - 0.1198
= 0.0802

We know that, 1 mole of water ≡ 18 g of water
Therefore, 0.0802 mole of water ≡ 1.4436 g of water.

Thus, 1.4436 g of water is excess in the experiment.