1) How much of a 0.567M hydrobromic acid solution would be needed to completely react with 523.4mL of a 1.267M mercury (I) nitrate solution?

2.    A reaction of 5.35g of solid methyl lithium (CH₃Li) reacts with 24.63g of solid carbamic acid (NH₂COOH) at 200K to produce methane gas (CH₄) and a dilithium carbamate salt (NHCO₂Li₂). If the completed reaction mixture heated to 350K in a 2.6L container, what is the pressure in the container?

 

3.    If the excess salt is dissolved in water, it will reform carbamic acid. At 350K, carbamic acid will decompose into NH₃ (g) and CO₂ (g). Assuming that the water vapor has no effect on the overall gas components or container volume, what is the final pressure in the container outlined in question 2 now?

 

4.    What are the partial pressures of each gas in the container? Provide two types of reactions (e.g. acid/base, oxidation/reduction, etc...) that will describe the reaction in question 2, and one additional to describe the reaction in 3 (this can be the same as one of the previous two answered for question 2, should it be applicable).

Answer:

I. 1169.57284mL of HBr is needed to completely react with 523.4mL of 1.267M of HgNO3

II. 7.146077955atm will be the final pressure in a system containing the products and the excessive reactant

III. 10.25829088atm will be the final pressure composed of CH4, NHCOOLi2, NH3, and CO2

IV.

CH4 has a partial pressure of 2.68924232atm

NH2CO2Li2 has a partial pressure of 1.344622709atm

CO2 and NH3 have partial pressures of 3.112212925atm

Reaction in Part II is Double-Displacement reaction (AB +CD-> AD +CB)

Reaction in Part III is Decomposition reaction (AB -> A+ B)

Step-by-step explanation

I.

The number of mols of HgNO3 can determine how much mols of HBr are needed to completely interact with HgNO3

This can be done by using the C1V1=C2V2 formula.

CxV= mol

(0.567M HBr) (xmL HBr) = (1.267M HgNO3) (523.4mL HgNO3)

This is equal to

xmL HBr= [(1.267M HgNO3) (523.4mL HgNO3)]/(0.567M HBr)

xmL HBr= 1169.57284mL

1169.57284mL of HBr is needed to completely react with 523.4mL of 1.267M of HgNO3

II.

This one is quite complex.

We know that we need pressure, this can be acquired with the Ideal Gas Law (PV= nRT) and (PV= (n1+n2+n3)RT for a mixture.

To get n1, n2 and n3, we have to use stoichiometry.

(n1= excess reactant, n2 and n3 are the products)

We can prepare the equation by knowing the molar masses of each substances first.

(If you don't know how, you can just comment, I will teach you; hint: You'll be needing a periodic table)

CH3Li = 21.975g/mol

NH2COOH= 61.044g/mol

CH4 = 16.042g/mol

NHCO2Li2= 72.91g/mol

This time, we have to know the whole reaction. We should balance it if needed

2 CH3Li + NH2CO2H -> 2 CH4 + NHCO2Li2

The next step is to determine the limiting reagent. How? Between CH3Li and NH2COOH, which has the lesser number of mols?

5.35g x (1mol/ 21.975g)x (1 NH2CO2H/ 2CH3Li) =0.1217292378mol

24.63g x (1mol/ 61.044g) x (2 CH3Li /1 NH2CO2H)= 0.8069589148mol

Which has the lesser number of mol? CH3Li. It is the limiting reagent

To determine the excess reagent, then we can subtract the number of limiting reagent which will react with the excess to the total number of excessive reactants.

24.63g x (1mol/61.044g) - 0.1217292378mol = 0.2817502196mol of NH2CO2H will be unreacted (this will be used in the ideal gas law)

*We will include the excessive reactant because there is no statement saying that only the products were left in the system*

Now, we have to determine the amount of products which will be formed

5.35g x (1mol/ 21.975g)x (2 CH4/ 2CH3Li)= 0.2434584755mol

5.35g x (1mol/ 21.975g)x (1 NHCO2Li2/ 2CH3Li)= 0.1217292378mol

To determine the pressure, we can apply the PV=(n1+n2+n3)RT rule. This is possible because this is an application of Dalton's Law of Partial Pressure.

P= [(n1+n2+n3)RT]/V

P=[(0.2434584755mol+0.1217292378mol+0.2817502196mol](0.082056Latm/molK)(350K)]/ 2.6L

P=7.146077955atm

7.146077955atm will be the final pressure in a system containing the products and the excessive reactant

III.

This is actually a continuation of part II.

We know that there is 0.2817502196mol of NH2CO2H left unreacted.

It is said to decompose into CO2 and NH3

We have to know the reaction first

1 NH2CO2H -> 1 CO2 + 1 NH3 (I didn't use H2O as a reactant because it act as a solvent)

Using the number of mol of the excess reagent, we can determine how many products can be formed

0.2817502196mol NH2CO2H x (1 CO2/ 1 NH2CO2H)= 0.2817502196mol CO2

0.2817502196mol NH2CO2H x (1 NH3/ 1 NH2CO2H)= 0.2817502196mol NH3

Instead of using P= [(n1+n2+n3)RT]/V, we can use P=[(n1+n2+n3+n4)RT]/V

P=[(0.2434584755mol+0.1217292378mol+0.2817502196mol+0.2817502196mol](0.082056Latm/molK)(350K)]/ 2.6L

P= 10.25829088atm

10.25829088atm will be the final pressure composed of CH4, NHCOOLi2, NH3, and CO2

IV.

This is simple.

We know the total pressure. We also know that there is a direct relationship between pressure and number of moles.

If we can get the mole fraction of each substance, then we can also get the partial pressure using the total pressure.

Formula

(mol of gas/ mole of mixture) = (partial pressure of gas/ total pressure)

Thus,

Total pressure x (mol of gas/ mol of mixture) = partial pressure of gas)

CO2 and NH3 (They have the same solution because they have the same amount of moles.)

[(0.2817502196mol/0.2434584755mol+0.1217292378mol+0.2817502196mol+0.2817502196mol)] x 10.25829088atm

CO2 and NH3 have partial pressures of 3.112212925atm

NH2CO2Li2

[(0.1217292378mol/0.2434584755mol+0.1217292378mol+0.2817502196mol+0.2817502196mol)] x 10.25829088atm

NH2CO2Li2 has a partial pressure of 1.344622709atm

CH4

[(0.2434584755mol/0.2434584755mol+0.1217292378mol+0.2817502196mol+0.2817502196mol)] x 10.25829088atm

CH4 has a partial pressure of 2.68924232atm

How do we know if we are right? We add the partial pressure.

2.68924232atm+1.344622709atm+ (3.112212925atm x2)=10.25829088atm

Is it the same as the final pressure? Yes? Then we are correct.