Question

What's the mass of the excess reactant?

10g Cs2CO3 reacts with 10g CaCl2
What mass of excess reactant should be left over?

Answer 1

m=6.6g

Explanation:

So first off, we should find a balanced chemical equation for this reaction.
`Cs_2CO_3+CaCl_2=>CsCl_2+CaCO_3`
From this, we see that `Cs_2CO_3` and `CaCl_2` combine in a 1-to-1 ratio.

The next step is to find the number of moles of each reactant. We use the formula `n=m/M`, where n is the number of moles, m is the mass in grams and M is the molar mass in g/mol.

`n CsCO_3 = m/M = (10g)/(325.82 g mol^-1)=0.03069mol`

`n CaCl_2 = m/M = (10g)/(110.99 g mol^-1)=0.09010 mol`

The excess amount of calcium chloride is equal to the initial moles of calcium chloride minus the moles of cesium carbonate = 0.05941 mol. (this works because of the 1-to-1 ratio explained above, so if all of the cesium carbonate in used up during the reaction, that's the same as the amount of calcium chloride used up).

Now we convert this back to mass using a rearranged version of the previous equation, `m=n*M=0.09010mol*110.99 g/(mol)=6.6g` (rounded to 2 significant figures)