Which of these is the largest quantity?

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

<u>Donde</u>:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

3 years ago

(Direct answer=Upvote cause I aint smart xD)

Vedmedyk [2.9K]

Paper is shredded, the paper is physically being shredded, iron rusting is due to oxygen reacting with it, gas is burned releases and odor which is a sign of chemical, shirt being bleached is chemical reaction.

5 0

3 years ago

he heat of fusion of tetrahydrofuran is . Calculate the change in entropy when of tetrahydrofuran melts at . Be sure your answer

lana [24]

Answer:

$\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Explanation:

Hello.

In this case, given the heat of fusion of THF to be 8.5 kJ/mol and freezing at -108.5 °C, for the required mass of 5.9 g, we can compute the entropy as:

$\Delta S=\frac{n*\Delta H}{T}$

Whereas n accounts for the moles which are computed below:

$n=5.9g*\frac{1mol}{72g} =0.082mol$

Thus, the entropy turns out:

$\Delta S=\frac{0.0819mol*8.5 kJ/mol}{(-108.5+273.15)K}\\\\\Delta S=1.8x10^{-3}\frac{kJ}{K}=1.8\frac{J}{K}$

Best regards.

3 0

3 years ago

What is oxidized in a galvanic cell with aluminum and gold electrodes ?

k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

$Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V$

$Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive): $Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V$

Notice that the overall cell potential upon summing is:

$E_{cell}=1.66 V + 1.50 V=3.16 V$

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0

3 years ago

Choose the atom with:

nata0808 [166]

Answer:

a )Li

b)O

c)F

Explanation:

a) Li-1s^2 2s^1

F-1s^2 2s^2 2p^5

it is easy to pull out e- from 2p orbit than 2s because 2s orbit is close to nucleus.Therefore Li have high ionisation enthalpy

b)oxygen ion is larger than Na because o have fewer proton

c)F because it requires only 1e to achieve stable noble gas configuration.Therefore to achieve stable nobke gas electonic configuration it accept 1e.

6 0

3 years ago