An unidentified metal M reacts with an unidentified halogen X to form a compound MX2. When heated, the compound decomposes by the reaction: When 1.12 g of MX2 is heated, 0.720 g of MX is obtained, along with 56.0 mL of X2 gas. Under the conditions used, 1.00 mol of the gas has a volume of 22.41 L. (a) What is the atomic weight and identity of the halogen X?

Video transcript

Hi everyone. This problem reads M X three is formed from the reaction between an unidentified metal M and an unidentified halogen X. The compound decomposes when heated and produces M. X and X two when 1.18 g of mX three is heated. 1.7 g of M and 34. mL of X two gas are obtained the volume of one mole of X two under the reaction condition is 22.41 Leaders determine the identity of halogen X and its atomic weight. Okay, so our mission here is to determine the identity identity of this halogen along with its atomic weight. So the way that we're going to do that is we're first going to figure out how many moles of X two we have and what we can start from is the volume of X two. That's given. We know we have 34. ml of X two. So our goal is to go from milliliters of X 22 moles of X two. So we're gonna do this by First looking at what's given in the problem that can help us to do this. And what's given in the problem is we know the moles of X two. So we know that we have one mole of X and we know we have the volume. Okay, 22.4 liters. So the volume of one mole of X two under the reaction is 22.4 liters. So let's go ahead and first convert this milliliters to liters. Okay, so in one leader there is mL. So our ml cancel and we're left in units of leaders. And now we can use the conversion that in one mole of X two. Okay. In one mole of X2 there is 22.41 leaders. Okay, so our leaders cancel and we're left with moles which is the unit that we wanted. So once we do this calculation we'll take 34.76 and divide it by 1000. And then divided again by 22.41. And we get 0. moles of X two. Alright, so we know our moles of X two. Now we're going to Figure out our mass. Okay, so our mass of X two is going to equal. What are we starting with? We're starting with We are starting with 1.18 g. Okay, so we have 1.18 g of MX three is heated. We need to subtract From, we need to subtract 1.07 g of em. Okay, so that's going to give us our massive X two because if we have 1.18 g of mx three is heated and we have 1.7 g of M that means these two numbers subtracted from each other is going to give us our massive X two. And when we do this calculation we get 0.11 g is our massive X two. So now what is our molar mass of X two. Okay. So remember molar masses grams per mole. So we know our mass of X two is 0. g per mole. We just calculated how many moles of X two we have. It's 0.1551. So let's plug that in. 0. moles. Okay, so we get 70.1 g per mole is the molar mass. Alright, we're almost done. So what is our atomic weight? Atomic weight of X. Which is what we're solving for is going to equal. So we know that we have we have in the problem 70.1. Okay, so we have 70.1. And we're gonna divide this by two. Because in the problem it's X two. So we only want the identity of X. So we're going to divide by two and we're going to get 35.35 g per mole. Now, if we look at the periodic table, what atomic weight is this? What element is this? The atomic weight for? Its the atomic weight for chlorine. Okay, so our final answer for this problem is the halogen X. Is chlorine with an atomic weight of 35. grams per mole. And that is our final answer. So we'll go ahead and highlight that here. And that is it for this problem? I hope this was helpful.