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A sample of metallic element X, weighing 4.316 g, combines with 0.4815 L of Cl2 gas (at normal pressure and 20.0oC) to form the metal chloride with the formula XCl. If the density of Cl2 gas under these conditions is 2.948 g/L, what is the mass of the chlorine?

also What is the atomic weight of X? What is the identity of X

2006-09-21 05:10:55 · 2 answers · asked by Anonymous in Science & Mathematics Chemistry

2 answers

Vol of Cl2 at 0 deg C = 0.4815*273/293 = 0.4486 l

So, 11.2 l of Cl will react with = 4.316*11.2/0.4486 = 107.756 g

X is Silver (Ag)

2006-09-21 05:27:18 · answer #1 · answered by ag_iitkgp 7 · 0 1

First:

(0.4815 L Cl2)*(2.948 g/L Cl2) = 1.419462 g Cl2

This means you have 2.838924 g Cl

Therefore you have 2.838924/35.453 (MW of Cl) = 0.08008 moles of Cl.

Then they join one to one, so you have 0.08008 moles of X. Which means X's MW = 4.316/0.08008 = 53.89 g/mol

This is closest to the element Manganese. (MW = 54.94)



Which means

2006-09-21 12:20:20 · answer #2 · answered by jimvalentinojr 6 · 0 0

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