Question

urea (CH4N2O mm=60.1 g/mol) is a common fertilizer than can be synthesized by the reaction of ammonia (NH3 mm=17.0 g/mol) with carbon dioxide (44.0 g/mol) according to the reaction below. a) 2NH3 (aq) +Co2 (aq)->CH4N2O (aq) +H2O (l)   in the synthesis of urea a chemist combines 130.4g of ammonia with 211.4 g of carbon dioxide and obtains 168.4 g of urea. b) theoretical yield of urea in grams c) percent yield for the reaction d) amount of excess reagent left over

Answer 1

Balanced equation is,

2 NH3 (aq.) + CO2 (aq.) ------------> CH4N2O (aq.) + H2O (l)

Moles of ammonia = mass / molar mass = 130.4 / 17.00 = 7.670 mol

Moles of carbon dioxide = 211.4 / 44 = 4.804 mol

From the balanced equation,

2 mol of NH3 needs 1 mol of CO2

hence, 7.670 mol of NH3 needs 7.670 / 2 = 3.835 mol of CO2 ( < 4.804 mol )

So, NH3 is limiting reagent.

From the balanced equation,

2 * 17.00 g. of NH3 forms 1 * 60.10 g. of urea

then, 130.4 g. of NH3 forms 130.4 * 60.10 / 34.00 = 230.5 g. of urea.

(b) Theoretical yield of urea = 230.5 g.

(c)

Percent yield = Actual yield * 100 / theoretical yield = 211.4 * 100 / 230.5 = 91.71 %

(d)

Moles of excess reagent remained = 4.804 - 3.835 = 0.969 mol of CO2

Mass of excess CO2 remained = 0.969 * 44 = 42.64 g.