Answer:
Final molarity of ammonium cation in the solution = 0.16 M
Explanation:
Complete Question
Suppose 2.59 g of ammonium nitrate is dissolved in 200. mL of a 0.40M aqueous solution of sodium chromate. Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.
Solution
2NH₄NO₃ + Na₂CrO₄ → (NH₄)₂CrO₄ + 2NaNO₃
We first convert the given parameters to number of moles
Number of moles = (Mass/Molar mass)
Molar mass of NH₄NO₃ = 80.043 g/mol
Number of moles of NH₄NO₃ = (2.59/80.043) = 0.03224 mole
Number of moles = (Concentration in mol/L) × (Volume in L)
Number of moles of Na₂CrO₄ = 0.4 × 0.2 = 0.08 Mole
2 moles of NH₄NO₃ react with 1 mole of Na₂CrO₄
So, it it evident that NH₄NO₃ is the limiting reagent as it is in short supply in the amount needed for the reaction.
So, the number of moles of ammonium ion in the product is also 0.03224 mole.
Molarity = (Number of moles)/(Volume L)
Molarity of ammonium ion = (0.03224/0.2) = 0.1612 mol/L = 0.16 M
Hope this Helps!!!
