Question

Suppose 0.153 g of iron(II) bromide is dissolved in 50. mL of a 10.0 m M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to 2 significant digits. ПМ x s ?

Answer 1

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Given moles of Febr₂ = Given man of febra 0.153gm molarmass 216 gm/noie = 7.08 x10-imel moles of AgNO3 = molarity * Volume (V) = 10x10m**0.050L = 5*10 mol Reaction fe Br2 + 2Ag Noz - Fe(NO3)2 + 2Agar from reaction it is clear that 10 1 mole febre reacts with zmole AgNO3 so 7.08 x 104 mole reacts with 2 x 7.08x10" - 14.16810" mole videos 1 Ay No3 so AgNO3 is a limiting reagent and 2 mol AgNO3 reacts with Imole feor d Camscanner'mol AgNO3 reads 5*10-4x1 = 2.5x104mol febr 2 mol si ob

moles of fe Bro remain in the solution = 7.088154-2-5x10 mole = 4-58*10* moli - Fe 2+ + 2or- FeBoz 4:58x104 4.58 x164 2x4.58 x 154 I Br-] = 2x4-58x10mor -0.01832 M. 0.05oL [Br] =10.018 IM Round to 2 significant digit Seanned with CamScanner