4.8g of calcium is added to 3.6 g of water. The following reaction occurs. Ca + 2H2O → Ca(OH)2 + H2

Question

(i) Which reagent is in excess? Explain your choice.

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

Answer 1

Ca(s) + 2H2O(l) ==> Ca(OH)2(s) + H2(g)

To find which reactant is limiting and which is in excess, we can calculate moles of each and divide that by the corresponding coefficient in the balanced equation. Whichever value is less is limiting and the other is in excess.

For Ca: 4.8 g Ca x 1 mol Ca/40.1 g = 0.120 moles Ca (÷1 -> 0.12) This is in excess

For H2O: 3.6 g H2O x 1 mol H2O/18 g = 0.20 moles H2O (÷2->0.10) This is limiting

So, Ca would be the reactant that is in excess

Now to find the mass of Ca left over at the end, we will find how much was used and subtract that from what we started with.

moles Ca used = 0.20 moles H2O x 1 mol Ca / 2 mol H2O = 0.10 moles Ca used

moles Ca left over = 0.12 moles - o.10 moles = 0.01 moles Ca left over

mass Ca left over = 0.01 moles Ca x 40 g Ca/mol Ca = 0.40 g Ca left over at end of experiment