6.55g of ammonia (NH3) react completely to the following reaction 2NH3+CO2-> CN2OH4+H2O what is the theoretical yield of urea (CN4OH4) for this reaction.if 8.75 mg urea recovers what’s percentage yiel

Answer 1

2 NH₃ + CO₂ --> CN₂OH₄ + H₂O

The above equation is already balanced. Starting with 6.55 g of ammonia,

6.55 g NH₃ (mol/17.04 g) ( mol urea/2 mol NH₃) (60.07 g/mol) = 11.5 g urea

% yield = experimental value (100) / theoretical value = 00.00875 g (100) / 11.5 g = 0.0758%

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