11.625 moles of Na₂O are produced.
<h3>What do you mean by the term mole ?</h3>
A mole is defined as the amount of a substance that contains exactly 6.02214076 × 10²³ ‘elementary entities’ of the given substance.
It can be represented by the following formula:
n = N/NA
To calculate moles of Na₂ O produced are -:
40 gm of sodium
3.00 gm of oxygen
4 Na+ O₂----->2Na₂O
First calculate limiting reagent,
Stochiometric ratio,
a/b=4/1
=4
Molar ratio,
M/n=4/22/3/22=4/22×32/3
=1.93
As a/b>m/n
So, sodium is limiting reagent
4 Na+ O₂----->2Na₂O
As 32 g of O₂ produced =124 g of Na₂O
SO,
3 gm of O₂ produced =124×3/32
=11.625 gm
11.625 gm of Na₂O can be produced .
Learn more about mole ,here:
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Answer: Actually, words designating the elements in the periodic table such as hydrogen, oxygen, and phosphorus are not Latin. They are Greek in Latin transliteration. ... Actually, words designating the elements in the periodic table such as hydrogen, oxygen, and phosphorus are not Latin. They are Greek in Latin transliteration.
Answer:
52.5 g
Explanation:
Density= mass/volume
Rearranging the equation gives us m= d(V)
m= d(V)
m= (10.5 grams/cm^3)(5 cm^3)
m= 52.5 g
Answer:
Explanation:
Hello,
In this case, for the given reaction at equilibrium:
We can write the law of mass action as:
![Keq=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
That in terms of the change 
due to the reaction extent we can write:
![Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7Bx%7D%7B%28%5BCO%5D_0-x%29%28%5BH_2%5D_0-2x%29%5E2%7D)
Nevertheless, for the carbon monoxide, we can directly compute as shown below:
![[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\](https://tex.z-dn.net/?f=%5BCO%5D_0%3D%5Cfrac%7B0.45mol%7D%7B1.00L%7D%3D0.45M%5C%5C)
![[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\](https://tex.z-dn.net/?f=%5BH_2%5D_0%3D%5Cfrac%7B0.57mol%7D%7B1.00L%7D%3D0.57M%5C%5C)
![[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\](https://tex.z-dn.net/?f=%5BCO%5D_%7Beq%7D%3D%5Cfrac%7B0.28mol%7D%7B1.00L%7D%3D0.28M%5C%5C)
![x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M](https://tex.z-dn.net/?f=x%3D%5BCO%5D_0-%5BCO%5D_%7Beq%7D%3D0.45M-0.28M%3D0.17M)
Finally, we can compute the equilibrium constant:
Best regards.
Answer: 0.39mol of the steam are in the cooker.
Explanation:Please see attachment for explanation
