Is 2CO + O2 -> 2CO2 A balanced or unbalanced equation

Iron reacts with oxygen to produce iron oxide (rust). This reaction is represented in an equation as 4Fe + xO2 → 2Fe2O3. Identif

Mila [183]

The answer is x = 3 cause 4=Fe=4 so its balanced
3=O2=3

3 0

4 years ago

1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea

Bad White [126]

Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

H₂: 3 moles
N₂: 1 mole
NH₃: 2 moles

The molar mass of the compounds is:

H₂: 2 g/mole
N₂: 28 g/mole
NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

H₂: 3 moles ×2 g/mole= 6 grams
N₂: 1 mole ×28 g/mole= 28 grams
NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

$mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}$

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

$mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }$

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

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7 0

2 years ago

Complete the decay equation for N-16: <br><br>16 7 N--> __+__

DanielleElmas [232]

The decay of the isotope N-16 occurs according to the reaction equation; 16/7 N -----> γ + 16/7 N.

<h3>What is a radioactive decay?</h3>

The term radioactive decay has to do with the spontaneous disintegration of an unstable nucleus to yield radiation.

The decay of the N-16 nucleus releases gamma rays which a radioactive radiation according to the equation; 16/7 N -----> γ + 16/7 N.

Learn more about radioactive decay:brainly.com/question/1770619

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5 0

2 years ago

how do scientists apply emission spectra to investigate the chemical composition of stars. 1 page long essay.

BARSIC [14]

Scientists use a spectroscope to receive the image of a spectra of light coming from faraway stars. Spectra is sometimes considered to be the fingerprint of the elements, with each elements having different spectra.

Explanation:

They do this by examining the spectrum of light from a star. When an ingredient is inspired, it gives off the light in distinct wavelengths, or colors.

The spectra of the Sun and stars presented bright and dark lines called Fraunhofer lines. Those were shown to be generated by elements transmitting or receiving light at specific wavelengths. Because each element emits or absorbs light only at specific wavelengths, the chemical distribution of stars can be resolved.

6 0

4 years ago

Read 2 more answers

A 50.0-ml sample of 0.200 m sodium hydroxide is titrated with 0.200 m nitric acid. calculate the ph in the titration after you a

attashe74 [19]

Hey there!:

Concentration of NaOH = 0.200 M

Concentration of HNO₃= 0.200 M

Total volume = 50.0 mL + 60.0 mL = 110 mL=> 0.11 L

The neutralization reaction between NaOH and HNO3 :

OH⁻ + H⁺ ----------> H₂O

So :

n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL => 0.012 moles of H⁺

n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻

Hence OH⁻ is limiting reagent .

Remaining moles of H⁺ = 0.012 - 0.01 => 0.002 moles

Concentration of H⁺ = 0.002 M / 0.11 L

Concentration of H⁺ = 0.01818 moles/L

Therefore:

pH = - log [ H⁺ ]

pH = - log [ 0.01818 ]

pH = 1.74

Hope that helps!

7 0

3 years ago