balancing equations and stoichiometry A compound called \'urea\', with the formu

Question

balancing equations and stoichiometry

A compound called 'urea', with the formula of (NH2)2Co, is commonly used in the production of fertilizer and explosives. It is prepared by the reaction of ammonia gas and carbon monoxide gas according to the following 2NH3 + CO rightarrow (NH2)2CO + H2O If 632.7 grams of ammonia is allowed to react with 1,142 grams of carbon monoxide, then please determine the answer to each of the following questions: Which of the two is reactants is the 'limiting reactant'? How much (NH2)2CO can be produced from the reaction? How much excess reactant (in grams) will be left over from the reaction?

Explanation / Answer

18

a) M.W of Ammonia is 17.03 g/mol

M.W of CO is 28.01 g/mol

632.7 g of ammonia will be 632.7/17.03 = 37.15 moles

1142 g of carbon monoxide = 40.77 moles

As per the equation 2 moles of ammonia and 1 moles of CO react to form one mole of urea

so the limiting reagent in this case as per the weight s taken is Ammonia

b) 2 moles of ammonia will produce one mole of urea , M.W of urea is 60.06 g/mol

We have 37.15 moles of ammonia so we will produce 37.15/2 = 18.57 moles of urea

which is 18.57 x 60.06 = 1115.6 g of urea will be produced

c) During the reaction all the ammonia will be consumed along with it 18.57 moles of CO will be consumed (because 2 moles of ammonia consumes 1 mole of CO)

18.57 moles of CO is 520.14 g of CO.

So remaining Co after the reaction will be 1142 - 520.14 = 621.8 g of CO will be left after the reaction

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