A 6.53-g sample of a mixture of magnesium carbonate
and calcium c...

Question

A 6.53-g sample of a mixture of magnesium carbonate
and calcium carbonate is treated with excess
hydrochloric acid. The resulting reaction produces 1.72 L
of carbon dioxide gas at 28 °C and 99.06 kPa pressure.
(c) Assuming that the reactions are complete, calculate
the percentage by mass of magnesium carbonate in the
mixture.

Accepted Answer

Hi everyone for this problem, we're told that magnesium oxide and strontium oxide react with carbon dioxide gas to form the following products. A mixture of magnesium oxide and strontium oxide, weighing 5.30 g was placed in a 2. liter Chamber containing c. 0. 2 gas With a pressure of 95. Kill a Pascal's at 21°C. After the reaction reached completion, the remaining carbon dioxide in the container has a pressure of 22.6 killer Pascal's determine the mass percentage of strontium oxide in the sample. So we want mass percent of strontium oxide in the sample. So for this problem, we have two reactions taking place. So let's go ahead and write those out first. Our reactions are magnesium oxide plus carbon dioxide give us this product and then we have strontium oxide plus carbon dioxide. Give us this product. So these are our two reactions. And because we're dealing with gasses here, we're going to need to use the ideal gas law and the ideal gas law tells us that P. V. Is equal to N. R. T. Okay. And so we're going to want to calculate the moles of carbon carbon dioxide that is reacting. And so in order to solve for moles, we're going to isolate our moles by dividing both sides of our equation by R. T. So that tells us that N. Is going to equal P. V. Over R. T. And this is what we're going to use to solve for our moles of carbon dioxide. So let's go ahead and plug in what we know based off of what we're given. Okay. And so p represents our pressure and in the problem we're told that our pressure is equal to we have a initial pressure and a final pressure. So we have 95 killer pascal's minus 22.6. Kill a pascal's gives us 72.4 killer pascal's. But we're going to need to convert this to A. T. M. Because r gas constant. R has the unit of A. T. M. So when we convert this we have 0.4 killer pascal's And one killer pascal There is pascal's and in one A. T. M. There's pascal's. So making sure our units cancel our killer pascal's cancel, our pascal's cancel. And we're left with unit of a T. M. So this gives us a final Pressure of a final answer for our pressure to be 0.7 one 45 A. T. M. So now we have our pressure to plug in next is volume. They tell us that The volume is equal to 2.50 L. Our our is our universal gas constant. This is not going to be given. It's a value that we should know and have memorized. It is 0.08 two 06 leaders per Atmos leaders times atmosphere over more times kelvin and T represents temperature. And so We're told it is at 21°C but our constant has our temperature in Kelvin. So we need to convert this temperature and°C to Kelvin. And for us to do that we add 273.15. So that gives us a temperature of 294. Calvin. Okay, so now we have everything that we need to plug in to solve for moles of CO2 reacted. So let's go ahead and do that. So we have N is equal to our pressure 0. A T M times are volume 2.50 l divided by R Which is 0.0 eight 206 Leaders times atmosphere over moles times kelvin. And our temperature is .15 Kelvin. We'll plug that in. And when we do that we get our moles is equal to 0. moles Of CO two. Okay, so now that we have our moles of CO2, I'm going to erase what we wrote down as are given for some room here. So now that we have our moles of CO2 reacted. What we can do next in order to solve for our mass percent of strontium oxide is well right out that are moles of magnesium oxide plus our moles of strong tm oxide equals our total moles of CO two reacted. Okay. And this is because we have a mixture. And so when we add up our components, we can solve for our mass percent. So we can say that our moles of m magnesium oxide is going to be X plus are moles of strontium oxide. We can say is why equals our total moles of seal to reacted. Okay. And from here we want to solve for why And we know that our total weight of our sample based off the problem is five point grams. So, we want to solve for why? Okay, when we saw for why? That will lead us to solving for our mass percent. So, when we when we rearrange this, why is going to equal 5. g minus All Right. So four X. We can say that our mass of magnesium oxide divided our massive magnesium oxide, divided by its molar mass, Is going to give us our moles of magnesium oxide. So 40 .305g per mole is going to give us X over 40.305g. So, this represents our moles of M G O. And then are mass of strong CME oxide Over its smaller mass, which is 103.62 grams per mole is going to represent 5. g - X. Based off of our equation above over its smaller mass. Okay, so this equals our moles of sean tm oxide. So when we plug this into our equation our X plus Y equation we get X over four .305g plus 5.30 - X. Over .62g equals Are moles of CO two reacted. Okay. Which we solved in the first part. So this is equal to 0. zero most. And we know that based off of the problem that our X is going to equal 1.51 g. Okay, so based off of that, this is our mass of magnesium oxide. So that means that our mass of strong titanium oxide if we erase this because we've already used it. We said that why is equal to 5.30 g -X. So we saw we know that X is 1.51 g. And so that means that our mass of strontium oxide Is going to be 5.30 g -X. So that makes our mass 3.79 g. So the last thing we can do now to solve for the mass percentage of strong strong strontium oxide is divide that by the total The weight of the mixture. So we have 3.79g divided by 5.30 g. Get times 100 because we're looking for mass percent is equal to 71 0.5% strong tm oxide. Okay, this is our final answer. And the end of this problem, I hope this was helpful.

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