Problem

Gaseous hydrogen has a density of 0.0899 g/L at 0 °C, and gaseous chlorine has a density of 3.214 g/L at the same tem-perature. How many liters of each would you need if you wanted 1.0078 g of hydrogen and 35.45 g of chlorine?

Relevant Solution
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Hi everyone. This problem reads at zero degrees Celsius. The densities of oxygen and nitrogen are 1.429 g per liter and 1.2506 g per liter respectively, calculate the volume of 15.999 g of oxygen in 14.6 g of nitrogen. Okay, so our goal here is to calculate the volume of the oxygen and nitrogen. Okay. And at the top were given the densities of each. Alright, so let's go ahead and write out the equation for density, density is equal to mass over volume. And our goal here is to calculate the volume of oxygen and nitrogen given the mass. Okay, so let's go ahead and start off with our oxygen. Okay, for oxygen were given the mass. So let's start with what we're given. Were given 15. g of oxygen. Alright, so now what we want to do is we want to go from grams of oxygen to volume of oxygen. And what we can use to do this Is the density that was given for oxygen. We can set this up as a conversion. Okay, so the density for the oxygen is 1.4-9 g per leader. So we need to set this up so that the g cancel each other. Okay, so there is and one leader There is 1. g of oxygen. Okay, so that's our density. So when we set it up this way we see our g cancel and we're left in units of leader, which is volume. That is what we want. So when we do this calculation, the volume that we're going to get for oxygen is 11.20 leaders. And this is our final answer for oxygen. So let's go ahead and write that down. We already specified that is oxygen. Okay, so it's 11.20 L is the final answer for that one. And let's go ahead and do the same thing, calculate the volume for nitrogen. We're going to start off with the mass that was given. The mass of nitrogen given is 14.006g. We want to go from grams of nitrogen to volume of nitrogen and again, we're going to use the density that was given and set it up as a conversion factor. Okay, so and one leader there is 1. g. This is the density that was given for the nitrogen. So we see here that the units of grams cancel and we're left with units of leaders which is a volume for our nitrogen. So once we do this calculation, We get a final answer of 11.199 leaders and this is our volume for the nitrogen. Okay, so that is it for this problem. We calculated the volume for each from the mass is given and we use the densities as the conversion factors. Okay, so that is the end of this problem. I hope this was helpful