Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate:H2PtCl6 + 2NH4+ = (NH4)2PtCl6 + 2H+The percipitate decomposes upon ignition, yielding metallic platinum and gaseous products:(NH4)2PtCl6 = Pt(s) + 2Cl2(g) + 2NH3(g) + 2HCl(g)What is the percentage of ammonium in a sample of 0.2213g of sample gave rise to 0.5881g of platinum?

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Asked Oct 6, 2019
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Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate:

H2PtCl6 + 2NH4+ = (NH4)2PtCl6 + 2H+

The percipitate decomposes upon ignition, yielding metallic platinum and gaseous products:

(NH4)2PtCl6 = Pt(s) + 2Cl2(g) + 2NH3(g) + 2HCl(g)

What is the percentage of ammonium in a sample of 0.2213g of sample gave rise to 0.5881g of platinum?

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Expert Answer

Step 1

When ammonium chloroplatinate decompose it produce ammonia gas, platinum, chlorine gas, and hydrochloric gas as shown in the reaction

The mass of platinum is 0.5881g, and mass of the sample is 0.2213 g.

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(NH)PtC Pt(s) +2C1, (g)+2NH(g)+2HClg) 4'

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Step 2

Since molecular mass of platinum is 195.08 g mol-1

No. of moles of platinum is equal to the given mass divided by molecular mass of platinum.

So the moles of platinum is 3.014 × 10-3.

As from above reaction

1 mole of p...

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0.5881 g no. of moles (n) 195.08 g/mol n 3.014x103 moles 2x3.014x10-3 6.028x103moles of ammonia

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