Mole ConceptHard

Question

If 0.250 g of an element M reacts with excess fluorine to produce 0.547 g of the hexafluoride, MF6, the element should be (Cr = 52, Mo = 95.94, S = 32, Te = 127.6, F = 19)

Options

A.Cr
B.Mo
C.S
D.Te

Solution

M + 3F2 → MF6

A g (A + 6 × 19)g

$\therefore 0.25\text{ g} \frac{A + 114}{A} \times 0.25 = 0.547 $$$\Rightarrow A = 95.96$$

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