Need help with limiting reactant problems in chemistry?

[匿名]

2017-11-29 3:03 pm

9.72g of copper(ii) nitrate is dissolved in 300. mL of a 0.50 M aqueous solution of sodium chromate. What is the final molarity of copper(ii) cation. The volume does not change.

I am super lost. Any help is greatly appreciated.

回答 (1)

戇戇居士

2017-11-29 3:26 pm

✔ 最佳答案

Molar mass of Cu(NO₃)₂ (copper(II) nitrate) = (63.5 + 14.0×2 + 16.0×6) g/mol = 187.5 g/mol

Initial number of moles of Cu(NO₃)₂ = (9.72 g) / (187.5 g/mol) = 0.05184 mol
Initial number of moles of Na₂CrO₄ = (0.50 mol/L) × (300/1000 L) = 0.15 mol

The balanced equation for the precipitation occurs :
Cu(NO₃)₂(aq) + Na₂CrO₄(aq) → CuCrO₄(s) + 2NaNO₃(aq)
Mole ratio Cu(NO₃)₂ : Na₂CrO₄ = 1 : 1

As (Initial number of moles of Cu(NO₃)₂) < (Initial number of moles of Na₂CrO₄),
Cu(NO₃)₂ is the limiting reactant, and thus almost all copper(II) cations are completely precipitated.
Hence, final molarity of copper(II) cations ≈ 0 M

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