All categories

3 years ago

How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.

Chemistry

1 answer:

MrRa [10]3 years ago

3 0

<h3>Answer:</h3>

78.34 g

<h3>Explanation:</h3>

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

<h3>Step 1: Writing the balanced equation for the reaction </h3>

The Balanced equation for the reaction is;

N₂(g) + 3H₂(g) → 2NH₃(g)

<h3>Step 2: Calculating the number of moles of NH₃</h3>

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ × 2

= 2.3 moles × 2

= 4.6 moles

<h3>Step 3: Calculating the mass of ammonia produced </h3>

Mass = Moles × molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles × 17.031 g/mol

= 78.3426 g

= 78.34 g

Thus, the mass of NH₃ produced is 78.34 g

You might be interested in

What is the reduction potential of a hydrogen electrode that is still at standard pressure, but has ph = 5.65 , relative to the

Zigmanuir [339]

Answer:

$\boxed{\text{-0.275 V}}$

Explanation:

We must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF} \ln Q$

1. Write the equation for the cell reaction

If you want the reduction potential, the pH 5.65 solution is the cathode, and the cell reaction is

Anode: H₂(1 bar) ⇌ 2H⁺(1 mol·L⁻¹) + 2e⁻; 0

Cathode: <u>2H⁺(pH 5.65) + 2e⁻ ⇌ H₂(1 bar)</u>; <u> 0 </u>

Overall: 2H⁺ (pH 5.65) ⇌ 2H⁺(1 mol·L⁻¹); 0

Step 2. Calculate E°

(a) Data

E° = 0

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C

n = 2

F = 96 485 C/mol

pH = 5.65

Calculations:

T = 25 + 273.15 = 298.15 K

$\text{H}^{+} = 10^{\text{-pH}} = 2.24 \times 10^{-5}\text{ mol/\L}\\\\Q = \dfrac{\text{[H}^{+}]_{\text{prod}}^{2}}{\text{[H}^{+}]_{\text{react}}^{2}} = \dfrac{(1.00)^{2}}{(2.24 \times 10^{-5})^{2}} =2.00 \times 10^{9}\\\\\\E = 0 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln{2.00 \times 10^{9}}\\\\= -0.01285 \times 21.41 = \textbf{-0.275 V}\\\text{The cell potential for the cell as written is }\boxed{\textbf{-0.275 V}}$

6 0

3 years ago

A Starting substance in a chemical reaction is called a _______________.

djverab [1.8K]

Answer:

1.)REACTANT

These starting substances of a chemical reaction are called the <em>reactants</em>, and the new substances that result are called the products.

5 0

3 years ago

You start with 1 L of CO2 at standard temperature and pressure in a closed container. If you raise the temperature of the gas, t

pychu [463]

Answer:

Increase

Explanation:

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

If the initial temperature and pressure is standard,

Pressure = 1 atm

Temperature = 273.15 K

then we increase the temperature to 400.0 K, The pressure will be,

1 atm / 273.15 K = P₂/400.0K

P₂ = 1 atm × 400.0 K / 273.15 K

P₂ = 400.0 atm. K /273.15 K

P₂ = 1.46 atm

Pressure is also increase from 1 atm to 1.46 atm.

8 0

2 years ago

How much heat (in Joules) is needed to raise the temperature of 257g of ethanol (cethanol=2.4 J/g°C) by 49.1°C?

Sonbull [250]

Answer:

Q = 30284.88 j

Explanation:

Given data:

Mass of ethanol = 257 g

Cp = 2.4 j/g.°C

Chnage in temperature = ΔT = 49.1°C

Heat required = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = 257 g× 2.4 j/g.°C × 49.1 °C

Q = 30284.88 j

8 0

3 years ago

Given the reaction: A + B <--> C + D

Lady_Fox [76]

Answer:

A.) 4.0

Explanation:

The general equilibrium expression looks like this:

$K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }$

In this expression,

-----> K = equilibrium constant

-----> uppercase letters = molarity

-----> lowercase letters = balanced equation coefficients

In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.

$K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }$ <----- Equilibrium expression

$K = \frac{[2 M] [2 M]}{[1 M] [1 M] }$ <----- Insert molarities

$K = \frac{4}{1 }$ <----- Multiply

$K = 4$ <----- Divide

6 0

2 years ago