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Suppose 12.7 g of iron(ii) bromide is dissolved in 250. ml of a 0.60 m aqueous solution of silver nitrate.rnrncalculate the final molarity of iron(ii) cation in the solution. You can assume the volume of the solution doesn't change when the iron(ii) bromide is dissolved in it.rnrnround your answer to 3 significant digits.
Accepted Answer
Balanced chemical equation:
FeBr2(aq) + 2AgNO3(aq) --→ Fe(NO3)2(aq) + 2AgBr(s)
Moles of FeBr2:
moles FeBr2 = mass / molar mass
moles FeBr2 = 12.7 g / 215.86 g/mol
moles FeBr2 = 0.0588 mol
Moles of Fe2+ ions:
moles Fe2+ = moles FeBr2 = 0.0588 mol
Total volume of the solution:
Total volume = volume of FeBr2 solution + volume of AgNO3 solution
Total volume = 250. mL + 0 mL = 250. mL = 0.250 L
Final molarity of Fe2+ ions:
Molarity Fe2+ = moles Fe2+ / total volume
Molarity Fe2+ = 0.0588 mol / 0.250 L
Molarity Fe2+ = 0.235 M
Rounded to 3 significant digits:
Molarity Fe2+ = 0.24 M
FeBr2(aq) + 2AgNO3(aq) --→ Fe(NO3)2(aq) + 2AgBr(s)
Moles of FeBr2:
moles FeBr2 = mass / molar mass
moles FeBr2 = 12.7 g / 215.86 g/mol
moles FeBr2 = 0.0588 mol
Moles of Fe2+ ions:
moles Fe2+ = moles FeBr2 = 0.0588 mol
Total volume of the solution:
Total volume = volume of FeBr2 solution + volume of AgNO3 solution
Total volume = 250. mL + 0 mL = 250. mL = 0.250 L
Final molarity of Fe2+ ions:
Molarity Fe2+ = moles Fe2+ / total volume
Molarity Fe2+ = 0.0588 mol / 0.250 L
Molarity Fe2+ = 0.235 M
Rounded to 3 significant digits:
Molarity Fe2+ = 0.24 M
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