How Many Anions Are There In 2.50 G Of MgBr2?


2 Answers

John McCann Profile
John McCann answered
2.50 grams MgBr2 (1 mole MfBr2/184.11 grams)(2 mole Br/1 mole MgBr2)(6.022 X 10^23/1 mole Br)

= 1.64 X 10^22 anions of bromine in 2.50 grams of magnesium bromide
Zhu Yugang Profile
Zhu Yugang answered
M(MgBr2)=184 g/mol,n=m/M(MgBr2)=2.5/184=0.0136 mol,the Anions are Br-,so n(Br-)=0.0136×2=0.0272 mol.There are 0.0272 mol Anions In 2.50 G Of MgBr2.
thanked the writer.
John McCann
John McCann commented
Question asked " how many ", not how many moles. Assuming your moles are correct, I will complete below.

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