(b) How many liters of CO2 gas were produced if the density of CO2 is 1.799 g/L?

Video transcript

Hello everyone today. We have the following problem. An unknown mass of a lead bicarbonate is allowed to react with 200 ml of .1253 molar of hydrogen sulfate. This forms a solution of lead sulfate and carbon dioxide gas which is removed via heat. The resulting solution is tight traded with 65. ml of .1344 moller of sodium hydroxide to completely neutralize the excess acid, calculate the volume of C. 02 gas formed by the initial reaction. If the density of C. 02 at 25 degrees Celsius is 1.808 g per liters. So the very first thing that we have to note is what does our reaction actually look like? And so our reaction, as noted by the question is going to be this lead bicarbonate, which is a solid reacting with our hydrogen sulfate in the acquis form. And this is going to form as in the question as per the question, lead sulfate in the acquis form as well as carbon dioxide in the gaseous form and water as a liquid. And so this from this we can derive what's known as a neutralization reaction. So we have a neutralization reaction. We're in we have our hydrogen sulfate in the acquis form here reacting with two moles of our sodium hydroxide that we used to titrate this with to give us sodium sulfate Aquarius and two moles of water in the liquid form. So now we can do some calculations. So first you want to find how many moles of our H two S. 04 we have from our initial solution. And so with that we're simply going to take the moles that we found our zero point 125, 3 moles of Rh two s 04. And we are going to divide it by our one leader. And since we used zero we used mL 200 mL is going to equal 2000.2 leaders. We simply just took the number of leaders and divided that by 1000. When our units cancel, we are left with 0. moles Of our H two S 04. No, we must find the moles of our excess H two S 04. So we found our initial amount of moles. And now we have to find out how much of excess that we have once has reacted with their in a our sodium hydroxide. So what we do As we take our polarity are 0.1344 moles of our sodium hydroxide. And of course that will be over one leader. Since that is a polarity. And we're going to multiply that By R 0.065- liter that we have from the question stem because we had our 65. middle leaders. And so we just divided that number by 1000 to get our leaders here in this problem. And we then lastly multiply that by a conversion factor a multiple ratio That for every one mole of our H two S 04, we get two moles of n a O H. And so we got that from our neutralization reaction. The two came from the to the coefficient in front of the sodium hydroxide and the one came from the coefficient that was in front of the hydrogen sulfate. And so when our units cancel out for this problem, We end up with four . times 10 to the negative third malls of our H two S 04 after it has reacted with sodium hydroxide. And so now we can therefore therefore find out how many moles reacted with our lead carbonate. And so we simply just subtract the two moles of our hydrogen sulfate that we found. So we're going to take our 0.2506 and subtract that by a 4.3814 times 10 to the negative third moles. And that's going to give us 0. moles of hydrogen sulfate. Now we can use that number and find out how many moles of CO2 was formed. So with our moles of CO2 that was formed, that will be that number that we just found. The number of moles are 0. moles of H two S 04 Times a conversion factor are multiple ratio being at one mole of CO gives us one mole of H two S Which was given to us by our coefficients. Once again, this one in front of the CO2 is that coefficient And the same with the H two S 04. Once our units cancel out, We get 0. malls of CO2. And so now for our final step of calculating the volume of CO2. So our volume of our carbon dioxide gas is going to be that 0.2068 moles of our carbon dioxide Multiplied by the molar mass of common dioxide. So we have one mole of co two and this is according to the periodic table, 44.01 g of CO two. We then have to get rid of the grams to get leaders. So we use the conversion factor per the periodic table. We use the density of CO2. We use one liter of C 02 equals 1.808 g of c 02. And once our units cancel out for this, we end up with 0.503 leaders of CO2 as our final answer. Overall, I hope this helped. And until next time