1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Serhud [2]
3 years ago
9

For this reaction, 11.5 g nitrogen monoxide reacts with 9.91 g oxygen gas. nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (

g) What is the maximum mass of nitrogen dioxide that can be formed? g What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete?
Chemistry
1 answer:
Artist 52 [7]3 years ago
8 0

Answer : The mass of nitrogen dioxide is, 17.6 grams

The formula of limiting reagent is, NO

The mass of excess reagent remains is, 3.74 grams

Explanation : Given,

Mass of NO = 11.5 g

Mass of O_2 = 9.91 g

Molar mass of NO = 30 g/mol

Molar mass of O_2 = 32 g/mol

First we have to calculate the moles of NO and O_2.

\text{Moles of }NO=\frac{\text{Given mass }NO}{\text{Molar mass }NO}

\text{Moles of }NO=\frac{11.5g}{30g/mol}=0.383mol

and,

\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}

\text{Moles of }O_2=\frac{9.91g}{32g/mol}=0.309mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

From the balanced reaction we conclude that

As, 2 mole of NO react with 1 mole of O_2

So, 0.383 moles of NO react with \frac{0.383}{2}=0.192 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and NO is a limiting reagent and it limits the formation of product.

Moles of excess reagent remains = 0.309 - 0.192 = 0.117 mol

Now we have to calculate the mass of excess reagent remains.

\text{ Mass of excess reagent}=\text{ Moles of excess reagent}\times \text{ Molar mass of excess reagent}

Molar mass of O_2 = 32 g/mole

\text{ Mass of excess reagent}=(0.117moles)\times (32g/mole)=3.74g

Now we have to calculate the moles of NO_2

From the reaction, we conclude that

As, 2 mole of NO react to give 2 mole of NO_2

So, 0.383 mole of NO react to give 0.383 mole of NO_2

Now we have to calculate the mass of NO_2

\text{ Mass of }NO_2=\text{ Moles of }NO_2\times \text{ Molar mass of }NO_2

Molar mass of NO_2 = 46 g/mole

\text{ Mass of }NO_2=(0.383moles)\times (46g/mole)=17.6g

You might be interested in
Which of the following is an example of chemical weathering?
REY [17]

Answer:

the answer is C acid dissolving limestone

Explanation:

6 0
2 years ago
Read 2 more answers
What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?
bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0
3 years ago
Read 2 more answers
Enter the molecular equation representing aqueous nitric acid and aqueous ammonia reacting. express your answer as a balanced mo
PtichkaEL [24]
Aqueous nitric acid and aqueous ammonia reacts to form ammonium nitrate 
 HNO₃(aq)+NH₃(aq) = NH₄NO₃(aq)
 HNO₃ +NH3 = NH₄ (+) + NO₃ (-)
Therefore the net ionic equation will be;
 H⁺(aq) + NH₃ = NH₄⁺ (aq)

4 0
3 years ago
Read 2 more answers
Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

6 0
3 years ago
How many grams are there in 2.3*10^24 atoms of silver nitrate ?​
Paul [167]
410g Ag

2.3*10^24 atoms

1 molcule Ag- 6.02g*10^3
6 0
3 years ago
Other questions:
  • How many atoms are in 16.67g of Calcium Nitrate?
    15·2 answers
  • 65 m^3 are fed / h of benzene to a reactor, it is requested: to. What is the mass flow fed in kg / h? b. And the molar flow in m
    15·1 answer
  • Given 8.25 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100
    10·1 answer
  • Complete the equation for the combustion of wood (assume carbon is the combustible material)
    8·1 answer
  • A chemist decomposes samples of several compounds; the masses of their constituent elements are listed. Calculate the empirical
    13·1 answer
  • Human impact on one system leads to changes in more of Earth's systems.
    14·2 answers
  • can someone make me a lab report? Please......... Objective(s): In your own words, what was the purpose of this lab? Hypothesis:
    14·2 answers
  • What's the answer to this and why? need a answer asap​
    5·1 answer
  • How many moles of KCIO3 are required to produce 58.3 grams of 02? <br> 2 KCIO3 —&gt; 2 KCI + 3 O2
    8·1 answer
  • What is the volume of a piece of metal that has a density of 6.68g.cm^3 and has a mass of 1975 g?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!