How many bromide ions are present in 0.55 g of iron(III)bromide? A) 1.1 × 10²¹ ions B) 3.4 × 10²¹ ions C) 3.3 × 10²³ ions D) 9.9 × 10²³ ions E) 2.9 × 10²⁶ ions

Question

Quizplus · Accepted Answer

First, we need to calculate the molar mass of iron(III) bromide: 
FeBr3 = 1 iron atom (55.85 g/mol) + 3 bromine atoms (79.90 g/mol) 
= 55.85 g/mol + 239.70 g/mol 
= 295.55 g/mol

Next, we need to calculate the number of moles of iron(III) bromide: 
moles = mass/molar mass 
moles = 0.55 g / 295.55 g/mol 
moles = 0.00186 mol

Finally, we can use the mole ratio between iron(III) bromide and bromide ions to calculate the number of bromide ions present: 
1 mol FeBr3 : 3 mol Br- 
0.00186 mol FeBr3 : x mol Br-

x = 0.00186 mol FeBr3 × 3 mol Br-/1 mol FeBr3 
x = 0.00558 mol Br-

To get the number of ions, we need to multiply by Avogadro's number: 
0.00558 mol Br- × 6.022×10^23 ions/mol 
= 3.35×10^21 ions

Therefore, the answer is B) 3.4 × 10^21 ions.

How Many Bromide Ions Are Present in 0