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×(C) MyLab and M × y ill Course Home or C: vellum.ecohege.com/course.html?courseld = 141 70229&OpenVellurnH; MAC-492a686a07f1a27b6fedc8a209541 fcS® 1 000 1 MasteringChemistry: Ch. 7: Chem Rxns & Quantities HW Googie Chrome Secure l https://session.masteringchemistry.com/myct/itemView?assignmentProblemID=91 177069 Ch. 7 Chem Rxns& Quantities HWQuestion 20 Question 20 Part A Urea (CH N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NHs) with carbon dioxide as follows 2NFh(agl t CO2(aq) CH,N20(aq) + H2O(1) In an industrial synthesis of urea a chemist combines 138.8 kg of ammonia with 211.4 kg of carbon dioxide and obtains 172.0 kg of urea Determine the limiting reactant Express your answer as a chemical formula. Submit My Answers Give Up Incorrect; Try Again; 6 attempts remaining Part B Determine the theoretical yield of urea

Explanation / Answer

The balance reaction is as follows:

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

Given that

NH3= 138.8 kg or 138800 g

CO2 = 211.4 kg or 211400 g

And

CH4N2O , Urea = 172.0 kg or 172000 g

First calculate the number of moles as follows:

Number of moles = amount in g/ molar mass

NH3: 138800 g /17.031 g/mol

= 8150 moles

CO2 : 211400 g /44.01 g/mol

= 4803.5 Moles

4803.5 Moles CO2 * 2 Mole NH3/1 mole CO2

= 9607 mole NH3

OR

8150 moles NH3 * 1 Mole CO2/2 mole NH3

= 4075 Mole CO2

Here CO2 is present in excess.

Excess CO2 = 4803.5 Moles-4075 Mole CO2

=728.5 Mole CO2

In this reaction the limiting agent is NH3 . The limiting agent has due to following properties:

8150 mole NH3 * 1 mole CH4N2O/ 2 mole NH3

= 4075 Moles CH4N2O

Amount of urea in g

4075 Moles * molar mass

=4075 Moles*60.06 g/mol

= 244744.5 g urea

= 244.75 Kg urea

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