To obtain a high yield of nitroacetanilide. three equivalents of the nitronium i

Question

To obtain a high yield of nitroacetanilide. three equivalents of the nitronium ion will be reacted with acetanilide in this experiment. If the molecular weight of acetanilide is 135.17 g/mol. what amount (mol) of nitronium ion is required? mol Using the equation from question 6: If the molecular weight of nitric acid is 63.01 g/mol and its density is 1.5129 g/mL. what volume (mL) of nitric acid is required? mL If the molecular weight of sulfuric acid is 98.08 g/mol and its density is 1.84 g/mL. what volume (mL) of sulfuric acid is required? mL Finish the following balanced equation showing the formation of the nitronium ion

Explanation / Answer

According to the problem; 1 mol acetinilide reacts with 3 mol NO2+

moles of acetinilide = 1 gm / 135.17 gm/mol = 0.0074 mol

  {C}{C}{C} 1 mol acetinilide reacts with 3 mol NO2+

0.0074 mol acetinilide reacts with 3*0.0074 mol NO2+ = 0.0222 mol NO2+

2. Molar volume of HNO3 required =

Density = mass/volume

Volume = mass/ density

Vm = 63.01 g/mol / 1.5129 g/mL = 41.65 mL/mol

3.

Molar volume of H3SO4 required =

Density = mass/volume

Volume = mass/ density

Vm = 98.08 g/mol / 1.84 g/mL = 53.3 mL/mol

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