Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of -164 °C. One possible strategy is to oxidize the methane to methanol, CH3OH, which has a boiling point of 65 °C and can therefore be shipped more readily. Suppose that 3.03 * 108 m3 of methane at atmospheric pressure and 25 °C is oxidized to methanol. What volume of methanol is formed if the density of CH3OH is 0.791 g>mL?

Video transcript

Hi everyone for this problem, we're told that nitric acid can be obtained from the reaction of nitrogen dioxide and water. The net reaction is shown below. Initially a 2.55 liter reaction vessel is filled with nitrogen dioxide gas at 1.5 A. T. M. And 28 degrees Celsius, calculate the volume of nitric acid produced from the reaction. Note that the density of nitric acid is 1.51 g per millimeter. So for this problem, we want to calculate the volume of nitric acid and something that we're going to need to remember in order to solve this problem is the ideal gas law, which tells us that P. V. Is equal to N. R. T. And what we can do here is we can, because we're given the density, we can calculate the volume using density. But first we need to solve for moles of and 02. And so we're going to go from moles of N 02, two mass of N 02. And then we can use our density to solve for our volume. Okay, so that's what's mapped out. So let's rearrange our ideal gas law so that we can solve for moles. Okay, so if we divide, I'm going to move this down. If we divide our ideal gas law by R. T. On both sides, we're going to get N. Which is our moles is equal to P. V over R. T. So that's where we're going to start. So let's start off by finding what is our pressure or plugging in what we are given. Okay, So we have our pressure we're told is 1.5 a T. M. And we're going to multiply that by B which is volume. And we're told that we have a 2.55 liter reaction vessel. Okay. And this is going to be over to move this over here. Our is our gas constant, which is zero point 08206 leaders per atmosphere over more times kelvin. And so this is a constant. So this is something that we should know. It wasn't given in the problem. And then our temperature because r universal gas constant is in kelvin and we're told that we have 28 degrees Celsius, we need to convert 28 degrees Celsius to kelvin and we can do that by adding so degrees Celsius plus 273.15 gives us 301.15 kelvin. So we needed to convert that first so we can plug that in 301.15 kelvin. So now we have everything that we need to solve four moles. So once we plug this in we get our mold is equal to zero point 1083 most as and 02. Okay, so now we have our moles of N L two and now we need to go to mass of N 02. So we can find our mass of N 02 from moles by using our multiple ratio and moller mass. So let's go ahead and do that. And so we have we just saw for our moles of N 02. So let's look at our reaction. We need to go from moles of N 02, two g of nitric acid. Because that's what we're looking for. We're looking for nitric acid. What is our multiple ratio of N 02 to nitric acid? When we look at our reaction, we're looking at N. 02 and nitric acid. And so we have to most of nitric acid for every three moles of N 02. And so are moles of N. O. To cancel. And we're going to be left with this should say most of H N 03. Let me rewrite that so we can see our units clearly. So two moles H and three. Okay, so two moles of nitric acid. So now we have moles of nitric acid but we want mass of nitric acid. So we need to look up the molar mass. And so in one mole of nitric acid using our periodic table, we see that our molar mass is 63 point 018 g of nitric acid. Okay, so let's look at our units. We see that moles of nitric acid cancel. And we're left with mass of nitric acid. So when we saw for this we get four, this is going to equal four point 549, nine g of nitric acid. So now that we have our mass of nitric acid, we can go to volume by using density. Okay, They were told that the density of nitric acid is 1.51 g per milliliter. We can say that we can use that unit conversion here and one millimeter. There is one point g of nitric acid and that's were that's from our density. And so when we solve that, that is going to give us our volume because our grams of nitric acid canceled and will be left with MIl leaders. So once we solve this we get 3. middle leaders is our volume. And this is our final answer. This is the volume of nitric acid produced from the reaction. That's the end this problem. I hope this was helpful.