Silver oxide, ammonium hydroxide, and ammonium nitrate form water and a compound containing 52.89% Ag, 20.61% N, 2.97% H, and 23.53% O. Calculate mass of each reactant to produce 45.35 g of compound.

Question

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Three compounds, silver oxide, ammonium hydroxide, and ammonium nitrate react to form a complex silver compound and water. The silver complex is found to contain 52.89% Ag, 20.61% N, 2.97% H, and 23.53% O. Calculate the mass of each reactant needed to produce 45.35 g of the complex silver compound.

Answer 1

The likely reaction would be

Ag₂O + 2NH₄OH + 2NH₄NO₃ ===> 2[Ag(NH₃)₂]NO₃ + 3H₂O

The product can also be written as AgH₆N₃O₃ which has a molar mass of 203.9 g/mole

45.35 g Ag(NH₃)₂]NO₃ x 1 mol Ag(NH₃)₂]NO₃ /203.9 g = 0.2224 moles Ag(NH₃)₂]NO₃

Ag2O: 0.2224 moles Ag(NH₃)₂]NO₃ x 1 mol Ag2O/2 mol Ag(NH₃)₂]NO₃ x 231.7 g/mol = 25.77 g Ag₂O

Half of the N comes from NH₄OH and half comes from NH₄NO₃. From this, we can do the following:

NH₄OH: 0.2224 moles Ag(NH₃)₂]NO₃ x 2 mol NH₄OH/2 mol Ag(NH₃)₂]NO₃ x 35.04 g/mol = 7.793 g

7.793 g/2 = 3.896 g NH₄OH

NH₄NO₃: 0.2224 mol Ag(NH₃)₂]NO₃ x 2 mol NH₄NO₃/2mol Ag(NH₃)₂]NO₃ x 80.04 g/mol = 17.80 g

17.80 g/2 = 8.900 g NH₄NO₃