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Chapter 4: Problem 109

A 1.00-g sample of an alkaline earth metal chloride is treated with excess silver nitrate. All of the chloride is recovered as 1.38 g of silver chloride. Identify the metal.

Short Answer

Expert verified
The alkaline earth metal in the chloride sample is Calcium (Ca).

Step by step solution

01

Find the moles of silver chloride and chloride ions

First, we need to determine the moles of silver chloride (AgCl) formed in the reaction. We are given the mass of silver chloride as 1.38 g, and we can find the molar mass of AgCl to be 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol (AgCl). Now, we can calculate the moles of AgCl formed: moles of AgCl = \( \frac{mass}{molar \:mass} \) moles of AgCl = \( \frac{1.38\: g}{143.32\: g/mol} \) = 0.00962 mol of AgCl Since AgCl dissociates into Ag and Cl ions in a 1:1 ratio, we therefore have exactly the same amount of chloride ions formed, that is 0.00962 mol of chloride ions.
02

Calculate the moles of alkaline earth metal chloride

We are given that the mass of the alkaline earth metal chloride is 1.00 g. Let MCl2 represent the metal chloride, where M is the alkaline earth metal and Cl2 is the chloride. The chloride content in MCl2 is the same as in the 1.38 g of AgCl, which means we have 0.00962 mol of chloride ions in the metal chloride sample. Since the ratio of M to Cl2 in MCl2 is 1:1, we therefore have 0.00962/2 = 0.00481 mol of the metal in MCl2.
03

Calculate the molar mass of the alkaline earth metal

Now we have the moles of the alkaline earth metal in MCl2, and we are given the mass of MCl2, 1.00 g. From this information, we can determine the molar mass of the alkaline earth metal: molar mass of M = \( \frac{mass \:of \:MCl2 - mass \:of \:Cl2}{moles \:of \:M} \) molar mass of M = \( \frac{1.00\: g - (0.00962 \:mol \times 35.45\: g/mol)}{0.00481\: mol} \) = 40.34 g/mol
04

Identify the alkaline earth metal

With the calculated molar mass of 40.34 g/mol, we can identify the alkaline earth metal by comparing it to the molar mass of known alkaline earth metals: 1. Be (Beryllium): 9.01 g/mol 2. Mg (Magnesium): 24.31 g/mol 3. Ca (Calcium): 40.08 g/mol 4. Sr (Strontium): 87.62 g/mol 5. Ba (Barium): 137.33 g/mol 6. Ra (Radium): 226.03 g/mol Comparing the obtained molar mass of the alkaline earth metal (40.34 g/mol) with the molar masses above, we can deduce that the metal is Calcium (Ca) as its molar mass is closest to our calculated value. Therefore, the alkaline earth metal in the chloride sample is Calcium.

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Most popular questions from this chapter

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