An element X forms an iodide 1XI32 and a chloride 1XCl32.
The iod...

Question

An element X forms an iodide 1XI32 and a chloride 1XCl32.
The iodide is quantitatively converted to the chloride when
it is heated in a stream of chlorine:
2 XI3 + 3 Cl2¡2 XCl3 + 3 I2
If 0.5000 g of XI3 is treated with chlorine, 0.2360 g of XCl3 is
obtained. (a) Calculate the atomic weight of the element X.

Accepted Answer

Hi everyone here we have a question telling us a medal from group to a forms of bromide with the formula M B R two. And chloride with the formula M C L. To the bromide can be converted into chloride. When reacted with chlorine gas. And it gives us the reaction M BR two plus chlorine form M C L two plus browning. When a 0.242554 g sample of M B R two is reacted with excess chlorine, 0.1321 g of M C L two is produced. And the question is, what is the atomic weight of the metal? M So because of the equation, we know that our Ratio of MBR two To MCL two Is 1 - one. So our moles of NBR two equals the mass of NBR two Over the formula mass of NBR two. Likewise, the moles of MCL two is going to equal the mess of MCL two over the formula mass By the MCL two. So our formula mass of MBR two equals M plus two times the molar mass of roaming because there's two of them, which is 79. g per mole. And likewise the formula mass of MCL two equals M the Mueller massive M plus two times the molar mass of chlorine, which is 35.45 to 7 graham's Permal. So the moles of NBR two Equals 0. Times in Plus two times 79.904. The moles of MCL two Equals 0.1321. I'm sorry, divided by, divided by M Plus two times 35.45 - seven. So they equal each other. So we have 0.2554, divided by M Plus 159. equals 0. over, M. 70.9054. So we end up with zero .2554. M plus 0. Times 70. Equals 0.1321. M plus 0.132, one times 159.808. So 0.1233, M Equals 3.00139764. Which means m equals 3.00139764, Divided by 0.1233. So m equals 24. grams per mole. And that is our final answer. Thank you for watching. Bye

Watch next