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3 years ago

If 7.45g of N2 and 36.5mL of 0.150M O2 react together, how many grams of N2O gas will be produced?

Chemistry

1 answer:

ira [324]3 years ago

4 0

Answer:

0.482 g of N₂O are produced.

Explanation:

The reaction is: 2N₂ + O₂ → 2N₂O

In order to produce 2 moles of N₂O, we need to make react 1 mol of oxygen and 2 moles of nitrogen.

We determine the moles of each reactant:

7.45 g / 28g/mol = 0.266 moles of nitrogen

36.5 mL . 0.15 M = 5.475 mmoles → . 1mol /1000 mmol = 0.00547 moles

Certainly the oxygen is the limiting reactant.

In order to determine the grams produced, we need to know the limiting reactant.

2 moles of N₂ need 1 mol of O₂

Then 0.266 moles of N₂ may react to (0.266 . 1) /2 = 0.133 moles

We only have 0.00547 moles, so there is not enough O₂.

1 mol of O₂ can produce 2 moles of N₂O

Then, the 0.00547 will produce (0.00547 . 2) /1 = 0.01095 moles

We convert to mass: 0.01095 mol . 44 g /mol = 0.482 g

That's the answer.

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PLEASE HELP FAST IM ABOUT TO FAIL PLEASE HELP PLEASE HELP FAST HELP HELP THIS IS DESPRATE

MariettaO [177]

1. 0.240 liters of water would be needed to dissolve 21.6 g of lithium nitrate to make a 1.3 M (molar) solution.

2. 2.9 M is the molarity of a solution made of 215.1 g of HCl is dissolved to make 2.0 L of solution.

3.83.3 ml of concentrated 18 M H2SO4 is needed to prepare 250.0 mL of a 6.0 M solution.

4. 135 ml of stock HBr will be required to dilute the solution.

5. 150 ml of water should be added to 50.0 mL of 12 M hydrochloric acid to make a 4.0 M solution

6. The pH of the resulting solution is 13.89

Explanation:

The formula used in solving the problems is

number of moles= $\frac{mass}{atomic mass of one mole}$ 1st equation

molarity = $\frac{number of moles}{volume}$ 2nd equation

Dilution formula

M1V1 = M2V2 3rd equation

1. Data given

mass of Lithium nitrate = 21.6 grams

atomic mass of on emole lithium nitrate = 68.946 gram/mole

Molarity is given as 1.3 M

VOLUME=?

Calculate the number of moles using equation 1

n = $\frac{21.6}{68.946}$

= 0.313 moles of lithium nitrate.

volume is calculated by applying equation 2.

volume = $\frac{0.313}{1.3}$

= 0.240 litres of water will be used.

2. Data given:

mass of HCl = 215.1 gram

atomic mass of HCl = 36.46 gram/mole

volume = 2 litres

molarity = ?

using equation 1 number of moles calculated

number of moles = $\frac{215.1}{36.46}$

number of moles of HCl = 5.899 moles

molarity is calculated by using equation 2

M = $\frac{5.899}{2}$

= 2.9 M is the molarity of the solution of 2 litre HCl.

3. data given:

molarity of H2SO4 = 18 M

Solution to be made 250 ml of 6 M

USING EQUATION 3

18 x V1= 250 x 6

V1 = 83.3 ml of concentrated 18 M H2SO4 will be required.

4. data given:

M1= 10M, V1 =?, M2= 3 ,V2= 450 ml

applying the equation 3

10 x VI = 3x 450

V1 = 135 ml of stock HBr will be required.

5. Data given:

V1 = 50 ml

M1= 12 M

V2=?

M2= 4

applying the equation 3

50 x 12 = 4 x v2

V2 = 150 ml.

6. data given:

HCl + NaOH ⇒ NaCl + H20

molarity of NaOH = 0.525 M

volume of NaOH = 25 ml

molarity of acid HCl= 75 ml

volume of HCl = 0.335 ml

pH=?

Number of moles of NaOH and HCl is calculated by using equation 1 and converting volume in litres

moles of NaOH = 0.0131

moles of HCl= 0.025 moles

The ratio of moles is 1:1 . To find the unreacted moles of acid and base which does not participated in neutralization so the difference of number of moles of acid minus number of moles of base is taken.

difference of moles = 0.0119 moles ( NaOH moles is more)

Molarity can be calculated by using equation 1 in (25 +75 ml) litre of solution

molarity = $\frac{0.0119}{0.1}$

= 0.11 M (pOH Concentration)

14 = pH + pOH

pH = 14 - 0.11

pH = 13.89

3 0

3 years ago

PLEASE HELP 2.0 x 10410 OTX 8.0 x 10

slava [35]

Answer:

$\frac{2.0.10(410)}{8.0.10( - 240)} = \frac{10(650)}{40}$

Explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

4 0

4 years ago

The equilibrium constant kp at 427°c for the reaction n 2( g) 3h 2( g) 2nh 3( g) is 9. 4 × 10 –5. what is δ g° for the reaction

SashulF [63]

ΔG° for the reaction is 5.47kJ mol⁻¹.

The energy that a substance has available for utilization in a chemical reaction or transformation is known as the Gibbs free energy. Things frequently change into other things that have less Gibbs free energy. The Gibbs free energy change indicates whether a chemical reaction will take place spontaneously or not.

By using the formula;

ΔG° = −RTlnKp

Where,

R = 8.3Jk⁻¹mol⁻¹

T = Temperature = 427 + 273 = 700 K

Kp = 8×10⁻⁵(given)

Substituting the value, we get,

ΔG° = −8.3 × 700 × ln(23×10⁻⁵)

ΔG° = −8.3 × 700 × (ln(2³)+ln 10⁻⁵)

= - 8.3 × 700 × (ln(2³)+ln 10⁻⁵)

= − 8.3 × 700 × (2.07−11.5)

=5.47×10⁴Jmol¹

=5.47kJ mol⁻¹

Therefore, ΔG° for the reaction is 5.47kJ mol⁻¹.

Learn more about Gibbs free energy here:

brainly.com/question/13765848

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7 0

2 years ago

A sample of 2.45g aluminum oxide decomposes into 1.3g of aluminum and 1.15g of oxygen. What is the percentage composition of the

Vitek1552 [10]

Answer:

% $Al = 53.1%$ %

% $O = 46.9$ %

Explanation:

If we know the grams of a chemical compound in a specific reaction, it is possible to know the percentage of each atom that composes it.

For the Aluminum Oxide in this problem, we know its total weight and the grams of each component.

therefore we can determine the percentage ratio of its components through:

For Al

% $Al = \frac{mass of Al}{mass of Aluminium oxide} . 100$ %

% $Al = \frac{1,3 g}{2,45 g} . 100$ %

% $Al = 53.1%$ %

In the same way for oxygen

% $O = \frac{mass of O}{mass of Aluminium oxide} . 100$ %

% $O = \frac{1,5 g}{2,45 g} . 100$ %

% $O = 46.9$ %

5 0

3 years ago

A 1.513 g sample of KHP (C8H5O4K) is dissolved in 50.0 mL of DI water. When the KHP solution was titrated with NaOH, 14.8 mL was

son4ous [18]

Answer:

0.501 M

Explanation:

KHP + NaOH → NaKP + H₂O

First we convert 1.513 g of KHP into moles, using its molar mass:

1.513 g ÷ 204.22 g/mol = 7.41x10⁻³ mol = 7.41 mmol

As 1 mol of KHP reacts with 1 mol of NaOH, in 14.8 mL of the NaOH solution there were 7.41 mmoles of NaOH.

With the above information in mind we can calculate the molarity of the NaOH solution:

7.41 mmol / 14.8 mL = 0.501 M

3 0

3 years ago