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Chapter 3: Problem 98

An element \(\mathrm{X}\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{XI}_{3}\) is treated with chlorine, $0.2360 \mathrm{~g}\( of \)\mathrm{XCl}_{3}$ is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

Short Answer

Expert verified
(a) The atomic weight of element X is approximately 30.975. (b) The element X is Gallium (Ga).

Step by step solution

01

Calculate the moles of XI3 and XCl3

First, let's calculate the moles of XI3 and XCl3 using their masses and molar masses. Moles of XI3 = (Mass of XI3) / (Molar mass of XI3) Moles of XCl3 = (Mass of XCl3) / (Molar mass of XCl3) We have masses of XI3 (0.5000 g) and XCl3 (0.2360 g), but we don't know the molar masses of these compounds because we don't yet know the atomic weight of X. Let's use the molar mass of Iodine(I) and Chlorine(Cl) for now. The molar mass of Iodine is 126.9045 g/mol and that of Chlorine is 35.453 g/mol. Molar mass of XI3 = (x + 3 * 126.9045) g/mol Molar mass of XCl3 = (x + 3 * 35.453) g/mol Using the mass values given, Moles of XI3 = 0.5000 / (x + 3 * 126.9045) Moles of XCl3 = 0.2360 / (x + 3 * 35.453)
02

Use stoichiometry of the reaction to relate moles of XI3 and XCl3

From the balanced equation, we see that 2 moles of XI3 react to form 2 moles of XCl3. So, their mole ratio is: Moles of XI3 : Moles of XCl3 = 1 : 1 Thus, Moles of XI3 = Moles of XCl3 Now, we can write: 0.5000 / (x + 3 * 126.9045) = 0.2360 / (x + 3 * 35.453)
03

Solve for the atomic weight of X

Let's solve for x in the above equation: 0.5000 * (x + 3 * 35.453) = 0.2360 * (x + 3 * 126.9045) Expanding and rearranging the terms: 0.5000x + 0.5000 * 3 * 35.453 = 0.2360x + 0.2360 * 3 * 126.9045 0.2640x = 0.2360 * 3 * 126.9045 - 0.5000 * 3 * 35.453 x ≈ 30.975 So, the atomic weight of element X is approximately 30.975.
04

Identify the element X

With the atomic weight of approximately 30.975, we can identify element X by checking the periodic table of elements. The element with an atomic weight close to 30.975 is Gallium (Ga), which has an atomic weight of 31. In this context, we can conclude that element X is Gallium (Ga). Answer: (a) The atomic weight of element X is approximately 30.975. (b) The element X is Gallium (Ga).

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Most popular questions from this chapter

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily HCl: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of HCl that can react with $0.500 \mathrm{~g}\( of \)\mathrm{Al}(\mathrm{OH})_{3}$ (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}$.

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2},\) and \(\mathrm{NaOH}\) : $2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)$ If the plant produces \(1.5 \times 10^{6} \mathrm{~kg}\) ( 1500 metric tons) of \(\mathrm{Cl}_{2}\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

(a) What is the mass, in grams, of \(2.50 \times 10^{-3} \mathrm{~mol}\) of ammonium phosphate? (b) How many moles of chloride ions are in \(0.2550 \mathrm{~g}\) of aluminum chloride? (c) What is the mass, in grams, of \(7.70 \times 10^{20}\) molecules of caffeine, \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\) (d) What is the molar mass of cholesterol if \(0.00105 \mathrm{~mol}\) has a mass of \(0.406 \mathrm{~g}\) ?
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