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suter [353]
3 years ago
11

Using 3 O2 molecules and 5 H2 molecules, how many water molecules can be produced? Do you have any left over?

Chemistry
2 answers:
makvit [3.9K]3 years ago
6 0

Answer:

5 molecules of H₂O can be produced

0.5 molecules of O₂ did not reacted

Explanation:

The reaction is:  2H₂(g)  +  O₂(g)  →  2H₂O (g)

Firstly we determine the limiting reactant:

2 moles of hydrogen need 1 mol of oxygen to react

We must know the moles of each.

6.02ₓ10²³ molecules is 1 mol

3 molecules are ____ 3 /6.02ₓ10²³ = 4.98×10⁻²⁴ moles O₂

5 molecules are ____ 5 / 6.02ₓ10²³ = 8.30×10⁻²⁴ moles H₂

2 moles of H₂ need 1 mol of O₂

Then 8.30×10⁻²⁴ moles of H₂ must need (8.30×10⁻²⁴ .1) / 2 = 4.15×10⁻²⁴ moles O₂. It is ok, because I have 4.98×10⁻²⁴ moles O₂. Oxygen is the reagent in excess, so the limiting is the H₂

1 moles of O₂ needs 2 moles of H₂ to react

Then, 4.98×10⁻²⁴ moles of O₂ must need (4.98×10⁻²⁴ .2) / 1 =9.96×10⁻²⁴ moles of H₂, we don't have enough H₂

So, in the reaction ratio is 2:2.

8.30×10⁻²⁴ moles of H₂ will produce 8.30×10⁻²⁴ moles H₂O

1 mol has 6.02×10²³ molecules

8.30×10⁻²⁴ must have (8.30×10⁻²⁴ . NA) = 5 molecules

The reagent in excess is the O₂. These means that there is oxygen that has not reacted.

We have 4.98×10⁻²⁴ moles O₂ and we used 4.15×10⁻²⁴ moles.

(4.98×10⁻²⁴ - 4.15×10⁻²⁴) = 0.83×10⁻²⁴ moles of oxgen hasn't reacted.

1 mol is contained by NA molecules

0.83×10⁻²⁴ moles are contained by (0.83×10⁻²⁴ . 6.02×10²³) = 0.5 molecules

Bond [772]3 years ago
4 0

Answer:

From 3 O2 molecules and 5 H2 molecules we can make 5 H2O molecules.

There remains 1 O-atom

Explanation:

Step 1: Data given

Number of O2 molecules = 3

Number of H2 molecules = 5

Step 2:

1 water molecule contains 1 H2 molecules and 1 O- atom

In 3 O2 molecules we have 6 O-atoms

In 5 H2 molecules we have 10 H-atoms

1 water molecules contains 2 H- atomes and 1 O-atom

From 3 O2 molecules and 5 H2 molecules we can make 5 H2O molecules.

There remains 1 O-atom

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Answer:

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Explanation:

1. F

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4 0
3 years ago
Formula los siguientes compuesto: Dietil eter, Etanol, Propanotriol, Acido Propanodioico, Pentanal, Pentano-2,4-diona, Metanoato
viva [34]

Answer:

Explanation:

En este caso para formular los compuestos, debes identificar el grupo funcional principal de la molecula. Una vez que eso está hecho, puedes intentar formularlo.

Empezaremos primero identificando el grupo funcional principal de la molécula, para luego formularlo correctamente.

Dietil eter: la terminación eter al final significa que pertenece al grupo de los éteres, el cual tiene como formula general R - O - R.

Etanol: debido a que termina en ol, este grupo pertenece a los alcoholes. Para formularlo solo se dibuja la molecula del etano, junto a un enlace con el grupo OH, como su formula general R - OH.

Propanotriol: igualmente termina en ol, por lo tanto es un alcohol, sin embargo, en este caso, tambien tiene la terminación prefija tri, asi que significa que hay 3 grupos OH en la molecula.

Acido propanodioico: esta es sencilla, porque empieza como acido, y solo hay un grupo funcional que empieza así y son los acidos carboxilicos, es decir, el grupo COOH (R - COOH) que es el carboxilo. Tiene el prefijo di, antes del oico, por lo que son dos carboxilos presentes en la molecula.

Pentanal: el sufijo al, significa que pertenece al grupo de los aldehidos, en este caso, posee el grupo carbonilo H - C = O.

Pentano - 2,4 - diona: la terminación ona significa que pertenece al grupo de las cetonas, (R - CO - R), parecido a los aldehidos, con la diferencia de que tiene grupos alquilos en lugar de un hidrogeno.

Metanoato de metilo: la terminación ato de ilo, pertenece a los esteres, (R - COOR) derivado de los acidos carboxilicos.

De aqui en adelante solo mencionaré los grupos funcionales pues ya se explicó el por que, por sus terminaciones:

Ciclohexano - 1.3 - diol: este pertenece a los alcoholes.

Acido heptanoico: acido carboxilico

Ciclobutil metil eter: eteres

Acetato de etilo: ester

2-metilbenzaldehído: aldehído unido a un grupo aromatico como el benceno.

Ciclohexanona: un ciclo (cadena cerrada) unido a un grupo carbonilo.

Butanona: cetona.

Observa la foto adjunta para que veas la formulación de cada una:

6 0
3 years ago
Match the following:
chubhunter [2.5K]

Answer:

1. a chemical reaction in which one substance breaks up into two or more new substances:  decomposition reaction

2. a reaction in which two or more substances combine to form a new substance: synthesis reaction

3. the reaction of an acid with a base to form a salt and water: neutralization reaction.

4. chemical compound formed when the negative ions from an acid combine with the positive ions of a base: salt

5. two ionic compounds reacting in solution to form two new compounds, one of which is insoluble: double displacement reaction.

6. a reaction in which an active metal displaces a less active metal or hydrogen from a compound solution (or a nonmetal replaces a nonmetal from a compound in solution): Single replacement reaction

Explanation:

1. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

Example: Li_2CO_3\rightarrow Li_2O+CO_2

2. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

Example: Li_2O+CO_2\rightarrow Li_2CO_3

3 and 4. Neutralization is a chemical reaction in which an acid and a base reacts to form salt and water. Salt is formed when cations or positive ions of base combine with anions or negative ions of acid.

Here KCN is the salt formed by combination of K^+ from base and CN^- from acid.

Example:  H^++CN^-+K^++OH^-\rightarrow K^++CN^-+H_2O

5. A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

Example: 2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)

6. Single replacement reaction is a chemical reaction in which more reactive element displaces the less reactive element from its salt solution.

Example: Zn+2HCl\rightarrow ZnCl_2+H_2

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