Which chemical equation describes an acid-base neutralization reaction?

In 1909 Fritz Haber discovered the workable conditions under which nitrogen, N2(g), and hydrogen, H2(g), would combine using to

labwork [276]

Answer : 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

Solution : Given,

Mass of $NH_3$ = 100 g

Molar mass of $NH_3$ = 27 g/mole

Molar mass of $N_2$ = 28 g/mole

First we have to calculate moles of $NH_3$ .

$\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}= \frac{100g}{27g/mole}=3.7moles$

The given balanced chemical reaction is,

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

From the given reaction, we conclude that

2 moles of $NH_3$ produced from 1 mole of $N_2$

3.7 moles of $NH_3$ produced from $\frac{1mole}{2mole}\times 3.7mole=1.85moles$ of $N_2$

Now we have to calculate the mass of $N_2$ .

Mass of $N_2$ = Moles of $N_2$ × Molar mass of $N_2$

Mass of $N_2$ = 1.85 mole × 28 g/mole = 51.8 g

Therefore, 51.8 g of nitrogen are needed to produce 100 grams of ammonia gas.

5 0

3 years ago

Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction?

Vlad1618 [11]

Answer: option B.

carbon + oxygen → carbon dioxide

Explanation:

8 0

3 years ago

Read 2 more answers

How many moles is 2.55 x 10 to the power of 26 atoms of Neon?

Yakvenalex [24]

Answer:

424 mol

Explanation:

Step 1: Given data

Number of atoms of Neon: 2.55 × 10²⁶ atoms

Step 2: Calculate the number of moles corresponding to 2.55 × 10²⁶ atoms of Neon

In order to convert atoms into moles, we need a conversion factor, which is Avogadro's number: there are 6.02 × 10²³ atoms of Neon in 1 mole of atoms of Neon.

2.55 × 10²⁶ atoms × (1 mol/6.02 × 10²³ atoms) = 424 mol

4 0

4 years ago

What is the mass of an object that has a density of 2.3 g/mL And a volume of 343ML

Bogdan [553]

Answer:

788.9g

Explanation:

density=m/v

2.3g/ml=m/343ml

=343×2.3

=788.9g

7 0

3 years ago

A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

3 years ago