Answer to Question #282091 in Organic Chemistry for daniel

Question

1.     The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C₆H₁₂O₆) to carbon dioxide (CO₂) and water (H₂O):

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

If 968 g of C₆H₁₂O₆ is consumed by a person over a certain period, what is the mass of CO₂ produced?                                                                                         

 

2.     Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide:

2NH₃(g) + CO₂(g) → (NH₂)₂CO(aq)+ H₂O(l)

In one process, 849.2 g of NH₃ are treated with 1223 g of CO₂.                       

(a) Which of the two reactants is the limiting reagent?

(b) Calculate the mass of (NH₂)₂CO formed.

 (c) How much excess reagent (in grams) is left at the end of the reaction?

 

 

Answer 1

$m(C_6H_{12} O_6) = 968g$

$M(C_6H_{12} O_6) = (12×6)+(1×12)+(16×6) =180g/mol$

$n(C_6H_{12} O_6)=\frac{m}{M} =\frac{968g}{180g/mol} =5.378mol$

$n(CO_2) =6×5.378 =32.268 mol$

$M(CO_2) =12+16×2 =44g/mol$

$m(CO_2)=n×M=32.268mol×44g/mol =1419.792g$

2.

$$n(NH3)=m/M=849.2g/17=49.953 mol$$ $$n(CO2)=1223g/44=27.795 mol$$

The limiting reagent is NH3. According to the reaction from 49.953 mol NH3 we can get 49.953/2=24.977 mol of (NH2)2CO.

$$m((NH2)2CO)=n×M=24.977×60=1498.62g$$

27.795 - 24.977 = 2.818 mol CO2 is left

at the end of the reaction.

$$m(CO2)=2.818×44=123.992 g$$