Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

I'm confused plz help! Chem!!?

Gold metal can be dissolved using sodium cyanide according to the reaction below:

4 Au(s) + 8 NaCN (aq) + O2(g) + 2 H2O(l) → 4 NaAu(CN)2(aq) + 4 NaOH(aq) 

 

If I added 86.7 g of Au to 750 mL of a 1.8 M solution of NaCN, which reagent would be limiting?

Type the reagent which is limiting (Au or NaCN or neither if there is no limiting reagent)

Limiting reagent:

Based on the limiting reagent, how many L of oxygen are required at 33 oC and 705 mmHg?

Give this answer to 3 significant figures.

O2 required:

1 Answer

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  • 1 month ago

    4 Au + 8 NaCN + O2 + 2 H2O → 4 NaAu(CN)2 + 4 NaOH

    (86.7 g Au) / (196.966569 g Au/mol) = 0.440176 mol Au

    (0.750 L) x (1.8 mol NaCN/L) = 1.35 mol NaCN

    0.440176 mole of Au would react completely with 0.440176 x (8/4) = 0.880352 mole of NaCN, but there is more NaCN present than that, so NaCN is in excess and

    Au is the limiting reagent.

    (0.440176 mol Au) x (1 mol O2 / 4 mol Au) = 0.110044 mol O2

    V = nRT / P = (0.110044 mol) x (62.36367 L mmHg/K mol) x (33 + 273) K /

    ( 705 mmHg) = 2.98 L O2

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