00:01
In this problem, we are combining 2 .56 grams of calcium carbonate with 250 milliliters of 0 .125 molar hydrochloric acid.
00:10
We would like to know if any calcium carbonate will remain once the reaction is complete, and we would also like to know what mass of calcium chloride can be produced.
00:20
Our first step in solving this problem is determining the limiting reactant.
00:24
So we're going to calculate our moles of our calcium carbonate and our moles of hydrochloric acid.
00:29
We can start by using the mass of calcium carbonate that we were given in the problem statement.
00:40
And we can divide this number by the molar mass of calcium carbonate, which is 100 ,08, 69 grams per mole.
00:55
And we find that there are 0 .0256 moles of calcium carbonate present.
01:05
Now we can do the same thing and find our moles of hydrochloric acid.
01:09
We know the molarity of our solution and the volume of our solution.
01:14
So we can take our molarity, that's per 1 liter.
01:25
And our solution is in terms of milliliters, so we can do a unit conversion.
01:33
And our solution has a volume of 250 milliliters.
01:39
And we end up with 0 .033moles of hydrochloric acid present...