What's the mass of the excess reactant?

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

1 Answer
Hasna K.
Nov 28, 2017

m=6.6g

Explanation:

So first off, we should find a balanced chemical equation for this reaction.
Cs2CO3+CaCl2⇒CsCl2+CaCO3
From this, we see that Cs2CO3 and CaCl2 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=mM, where n is the number of moles, m is the mass in grams and M is the molar mass in g/mol.

nCsCO3=mM=10g325.82gmol−1=0.03069mol

nCaCl2=mM=10g110.99gmol−1=0.09010mol

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⋅110.99gmol=6.6g (rounded to 2 significant figures)

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