Finding number of Anions given grams

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SUMMARY

The discussion focuses on calculating the number of anions in 1.50 g of magnesium bromide (MgBr2). The initial calculation used the formula 1.5 g MgBr2 x (1 mol MgBr2/184.3 g MgBr2) x (6.022 x 1023 anions/mol), resulting in 4.91 x 1021 anions. However, the correct answer is 9.81 x 1021 anions, indicating that the user failed to account for the stoichiometry of the compound, specifically the presence of two bromide ions (Br-) per formula unit of MgBr2.

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Homework Statement


How many anions are there in 1.50 g of MgBr2

Homework Equations

The Attempt at a Solution



1.5 g MgBr2 x (1 mol MgBR2/184.3 g MgBr2) x (6.022 x 1023 anions/mol)

which gave me answer of 4.91 x 1021

the correct answer is 9.81 x 1021

am i missing a step, or do i have to create a balanced equations? and use a mol to mol ratio?

any help is greatly appreciated.
 
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How many moles of Mg2+? How many moles of Br-?
 

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