Current time:0:00Total duration:10:18
0 energy points

Stoichiometry of a reaction in solution

Video transcript
What we want to do in this video is figure out the mass of calcium carbonate we need to react completely with 25 milliliters of 0.75 molar hydrochloric acid solution. And we have the reaction here. And just to remind ourselves what we're talking about we have this hydrochloric acid, and it is dissolved in water. It is an aqueous solution. The solvent is water, and this 0.75 molar tells us the molarity of that solution. It tells us the concentration or the moles of hydrochloric acid per liter of our entire solution. And we need to figure out the mass of this calcium carbonate that we need to react completely with the hydrochloric acid solution. This amount of it. Now before we do any stoichiometric problems, we have to make sure that our actual equation is balanced. So let's do that first. So on the left-hand side right here we've got a calcium. That's the only calcium. And on the right-hand side we only have 1 calcium. So everything looks good from the calcium point of view. On the left-hand side we have a carbon here. I'll do it in pink. We have 1 carbon here, and then we have no carbons here. And on the right-hand side we only have 1 carbon. Everything looks good so far. On the left-hand side we have 3 oxygens, no oxygens over here. And over here we have 1 oxygen and then 2 more. We have 3 on the right-hand side. Everything looks good so far. On the left-hand side we have 1 hydrogen right there. And then on the right-hand side we have 2 hydrogens. We have 2 hydrogens here. So to balance it let me put 2-- let me put a 2 in front of the hydrochloric acid, the aqueous solution of hydrochloric acid. Now the hydrogens balance out, and we haven't done the chlorines yet, which is good, because that would have messed up everything if they were already balanced. But let's see. Now we have 2 chlorines on the left-hand side. How many do we have on the right-hand side? We have 2. So now this equation is balanced after we threw that 2 coefficient on the hydrochloric acid. So now let's figure out what mass of calcium carbonate do we need to react completely with this much of 25 milliliters of a 0.75 molar hydrochloric acid solution. Now the whole way to think about it is I've given-- this is how this problem is different than everything else we've seen so far is that I haven't given you the mass of hydrochloric acid. I haven't told you the moles of hydrochloric acid. I've told you the amount of solution containing hydrochloric acid and the concentration. So what we're going to do is take this information to figure out the moles of hydrochloric acid in this reaction, how many moles do we have in this solution. And then for every 2 moles of hydrochloric acid we're going to need 1 mole of calcium carbonate. You're going to see that. And then we just have to figure out well, given that we're going to need half as many moles here as there are here, how much mass is that. So let's do that. So we are starting off with the 25 milliliters of-- let me write this-- 0.750 molar HCl solution, and let's convert this to moles of hydrochloric acid. So the first thing we need to think about is that molarity is always expressed in moles per liter. This is expressed in milliliters right now. So I want to get rid of the milliliters. I want it to be in liters. So I want to multiply it by liters-- I'll just write it as a capital L-- as liters per milliliter. And we know we have 1 liter per 1,000 milliliters, or there's 1/1000 of a liter per milliliters, however you want to view it. That's going to cancel out that milliliters. And then we want to convert-- we want to use this concentration information to figure out the actual number of moles of hydrochloric acid we're dealing with. So to do that we're going to multiply it times 0.750 moles of hydrochloric acid per liter of-- or I'll write this-- 0.750 molar hydrochloric acid solution. So let me make sure you understand what I'm doing here. I first want to get the 25 milliliters into liters, so I'm going to essentially divide by 1,000 and the units will work out. We have a milliliters in the numerator, milliliters in the denominator, we're left with a liter on top. And then I want to convert from the solution to the actual number of moles. And if something is a-- I could put a 0 right here-- 0.75 molar, that means that there is 0.75 moles of hydrochloric acid per liter of the solution. This is just-- I'm just saying you can call that the solution. And so this'll cancel out. We have the liters canceling out with the liters. And then you can even say the of 0.75 molar hydrochloric acid solution cancels out with the of .750 molar hydrochloric acid solution. And then you're left with-- let me stay in the same color-- 25 times 1 times 0.750 divided by 1,000 moles of hydrochloric acid. And let's get our calculator out to solve for this right here. So that is, let's see, 25 times 0.75 divided by 1,000 is equal to 0.01875. Let me write that down. So this is equal to 0.018-- if I want to stick with significant digits I really don't have any more than three significant digits here, so I really shouldn't have any more than three here. So I could say 0.018-- instead of 75 I'll just round it-- 8 moles of hydrochloric acid. And now we know that for every 2 moles of hydrochloric acid we need 1 mole of calcium carbonate. So we want to know the moles of calcium carbonate required. So let's do that. So let's multiply this times-- we want this thing to cancel out here. We want that to cancel out. So we want to know, so for every mole of hydrochloric acid, or for every 2 moles of hydrochloric acid, we see that there, for every 2 moles of hydrochloric acid we need 1 mole-- I'll do it in that same orange color-- we need 1 mole of calcium carbonate is required. And so these moles of hydrochloric acid is going to cancel out with that moles in the denominator. And we're essentially just going to take this number and multiply it by 1 and divide by 2, or just essentially divide it by 2. So let's do that. I could just take the original number right here, divided by 2-- I should only do it three significant digits-- so it's really-- well I'll just stay with this-- so 0, 0, 9, 3, 8 just to round it. So this is equal to, we're left with 0.00938 moles of calcium carbonate required. And we're not done yet. Remember, the question that we asked is what's the mass of calcium carbonate required. So we need to figure out, we need to essentially multiply this times the molar mass. So we want to convert this into grams. Let's convert this into grams. I'll switch colors just arbitrarily. And to put it into grams you have to figure out the molar mass of calcium carbonate. And the molar mass of calcium-- you could look up the atomic weight of calcium on a periodic table-- it is 40. And then you're going to have a carbon in there, that's 12, we've seen that a lot. Then you have an oxygen, which is 16, but we have three of them. So 16 times 3 is 48, so it's going to be 40 plus 12 plus 48, which is equal to 100. So it has a molar mass of 100. So times 100 grams of calcium carbonate for every-- let me just write that-- grams of calcium carbonate for every mole of calcium carbonate. And we know we're doing the right thing by multiplying by 100 because these units cancel out. Moles of calcium carbonate in the numerator, moles of calcium carbonate in the denominator. So those cancel out. So we essentially just have to multiply this thing times 100. So let's do that. So that thing times 100. So times 100 is equal to-- and we had rounded it-- so it's 0.938. So let me write that. 0.938. So it equals 0.938 grams of calcium carbonate.