Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate: H₂PtCl 6 + 2 NH4+* --> (NH4)2PtCl6 + 2H+ The precipitate decomposes on ignition, yeilding metallic platinum and gaseous products: (NH4)2PtCl6 --> Pt (s) + 2 Cl2 (g) + 2 NH3 (g) + 2 HCI (g) Calculate the percentage of ammonium in a sample if 0.2235-g gave rise to 0.8218 - g of platinum (At Mass = 195.08 g/mol). The molecular weight of ammonia is 17.0306 g/mol.
Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate: H₂PtCl 6 + 2 NH4+* --> (NH4)2PtCl6 + 2H+ The precipitate decomposes on ignition, yeilding metallic platinum and gaseous products: (NH4)2PtCl6 --> Pt (s) + 2 Cl2 (g) + 2 NH3 (g) + 2 HCI (g) Calculate the percentage of ammonium in a sample if 0.2235-g gave rise to 0.8218 - g of platinum (At Mass = 195.08 g/mol). The molecular weight of ammonia is 17.0306 g/mol.
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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 + 2 NH4* --> (NH4)2PtCl6 + 2H+
The precipitate decomposes on ignition, yeilding metallic platinum and gaseous products:
--> Pt (s) + 2 Cl2 (g) + 2 NH3 (g) + 2 HCI (g)
(NH4)2PtCl6 -
Calculate the percentage of ammonium in a sample if 0.2235-g gave rise to 0.8218 - g of platinum (At
Mass = 195.08 g/mol). The molecular weight of ammonia is 17.0306 g/mol.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0967fd87-d8ba-44d8-96d5-483b7f66c032%2F27ef726b-e0e4-46a1-82ab-a7cc6224ad9e%2F374ig1g_processed.png&w=3840&q=75)
Transcribed Image Text:Ammoniacal nitrogen can be determined by treatment of the sample with chloroplatinic acid; the
product is slightly soluble ammonium chloroplatinate:
H₂PtCl6 + 2 NH4* --> (NH4)2PtCl6 + 2H+
The precipitate decomposes on ignition, yeilding metallic platinum and gaseous products:
--> Pt (s) + 2 Cl2 (g) + 2 NH3 (g) + 2 HCI (g)
(NH4)2PtCl6 -
Calculate the percentage of ammonium in a sample if 0.2235-g gave rise to 0.8218 - g of platinum (At
Mass = 195.08 g/mol). The molecular weight of ammonia is 17.0306 g/mol.
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