The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so that the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is 2 H+ (aq) + Zn(s)¡ H2( g) + Zn2 + (aq). When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 °C is 0.951 L at a total pressure of 748 mmHg. What mass of hydrogen gas is collected?

Video transcript

Hi, everybody. Let's take a look at our next problem. It says the zinc in a copper plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots so that the hydrochloric acid can get to the zinc. The reaction between the acid and the zinc is two H plus aqueous plus zinc solid becomes H two gas plus ZN two plus aqueous. When the zinc in a certain penny dissolves the total volume of gas collected over water at 25.0 degrees. Celsius is 0.9 f 5 1 L at a total pressure of 748 millimeters of mercury. What mass of hydrogen gas is collected? So let's look at what information we have and what information we need. So we're looking for the mass of hydrogen that was produced, we have volume, we have temperature and we have pressure. So that means we can use our ideal gas Law equation PV equals NRT and SOL for N that would be the number of moles of hydrogen gas produced, which is a pretty easy conversion to change to grams. So let's solve this equation for N. So we have N equals PV divided by RT, but we need just a little bit of work to get to the values for pressure, volume and temperature that were going to use. So note that our zinc as it's being dissolved, this reaction is taking place in water and the gas that's collected is collected over water, which means that the gas produced will not just be hydrogen gas but also water vapor. So equals. So it's going to be hydrogen gas plus water and its gas form. So we need to isolate the pressure or the partial pressure of just the hydrogen gas. So we recall that our total pressure. So P total is the sum of the partial pressures of all the different gasses. So in this case, it would be P of H two plus P of plus the partial pressure of that water vapor, we have the total pressure given to us. So we're going to solve for the partial pressure of hydrogen. So ph two equals P total minus P of water. So we're given the total pressure of gas in our problem. It's a 748 millimeters of mercury and then subtract that partial pressure of water that we just look up as a standard amount. So when you look it up, it's expressed based on the temperature that partial pressure of water at 25 °C, which is the condition of our experiment is 23.8 millimeters of mercury. So we'll subtract that in our equation here, 23.8 millimeters of mercury and go ahead and solve for the partial pressure of hydrogen and we get 724 millimeters of mercury. However, we can't stop here because this isn't an SI unit. In order to put it in this equation with all these other units, we need it in the form of atmospheres. So we'll just put a conversion factor in here. There are 760 millimeters of mercury in one atmosphere. So the millimeters of mercury go on a denominator to cancel out. We have one atmosphere on the top and that's going to end up giving us 0.95 2632 atmospheres. We're now sort of summarizing what we have. I'm going to highlight in blue, we've gotten our pressure here, partial pressure of hydrogen that we just solved for in terms of volume. We are told that the total volume of gas collected over water is 0.951 L. We can use that even though it's a mixture of water vapor and the hydrogen gas. Since they take up the same amount of space, there won't be any added volume there. And the last bit of information we need is the temperature we're provided the temperature, but it's provided in degrees Celsius. And we need it in Kelvin for this equation for our units to cancel out. So our temperature is 25 °C and we add 273.15 to get to Kelvin, which gives us about 298 Kelvin. So we'll highlight that in blue because that is also a piece of the D or piece of the information that we need in our equation. And finally, R of course, is our gas constant. However, note that we need to use the version of R that has the unit liters atmospheres per mole Kelvin for everything to cancel out. So R value in that format is 0.08 206 liter atmosphere per mole Kelvin. So we'll highlight that in blue just to be consistent there. So we can just see right away where all our information is. So we're going to use our N equals PV divided by RT. So as we set our pressure here, 0.95 2632 atmospheres multiply that by our volume of 0.951 L. And then in the denominator, we have our R, 0.08206 leader atmospheres PM Calvin and then our temperature of 298 Kelvin. So let's look at how our units cancel out our atmospheres, cancel out our leaders cancel out. Kelvin cancels out, we're left with moles in the denominator, but those moles themselves are in the denominator of that R. So they'll end up being the units of our answer as we desire since we want the number of moles of hydrogen gas. So we plug all that into our calculator. We end up with 0.03 7047 moles and that's of hydrogen gas. But then in our final step, we need to convert that to grams since we are looking for the mass of hydrogen gas. So we need our molar or molecular mass of hydrogen gas. Don't forget that's H two. So we have our moles, we have one mole of hydrogen gas in a denominator of our conversion factor. And in the name reader, we have so we end up multiplying our moles of hydrogen gas by 2.016 g of hydrogen gas divided by one mole. And that gives us our final answer. We're looking for that. We end up with 0.0747 g of hydrogen gas. So that is our final answer. You see, it has three significant figures as we'd expect. And we look at our answer choices. Choice B matches this at 0.0747 g. See you in the next video.