a compound is decomposed In the laboratory and produces 1.40 g N and 0.20g H what is the empirical formula
1 answer:
First, find the number of moles for each element. The molar mass for nitrogen is 14 g/mol and that of hydrogen is 1 g/mol.
1.40 g N / 14 g/mol = 0.1 mol N
0.20 g H / 1 g/mol = 0.2 mol H
Find the mole ratio. Divide both numbers with the much lower value. In this case, it is 0.1 mol N.
For N: 0.1 ÷ 0.1 = 1
For H: 0.2÷0.1 = 2
Thus, the empirical formula is .
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I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .
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Step 3) Conversion of Dinitro toluene to trinitrotoluene.
This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.
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So over all reaction uses three reagents in order :
