The number of moles of Mg_3(PO_4)_2 containin | vedclass

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Question

(B) The chemical formula $Mg_3(PO_4)_2$ contains $8$ oxygen atoms per formula unit.Therefore,$1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen a

Vedclass · Accepted Answer

$3.125 	imes 10^{-2}$

Vedclass · Answer

(B) The chemical formula $Mg_3(PO_4)_2$ contains $8$ oxygen atoms per formula unit.Therefore,$1$ mole of $Mg_3(PO_4)_2$ contains $8$ moles of oxygen atoms.To find the moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms,we use the ratio:$	ext{Moles of } Mg_3(PO_4)_2 = \frac{	ext{Moles of oxygen atoms}}{8}$$	ext{Moles of } Mg_3(PO_4)_2 = \frac{0.25}{8} = 0.03125 = 3.125 	imes 10^{-2}$.

The number of moles of $Mg_3(PO_4)_2$ containing $0.25$ moles of oxygen atoms is:

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