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VMariaS [17]
4 years ago
12

2 H2 + O2 → 2 H2O How many grams of H2O are produced when 2.50 moles of oxygen are used?

Chemistry
1 answer:
vodomira [7]4 years ago
8 0

Answer:

90g

Explanation:

The equation for the reaction is given below:

2H2 + O2 → 2H2O

From the equation above,

1 mole of O2 produced 2 moles of H2O.

Therefore, 2.5 moles of O2 will produce = 2.5 x 2 = 5 moles of H2O.

Now let us convert 5 moles to H2O to grams to obtain the desired result. This is shown below:

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Number of mole of H2O = 5 moles

Mass of H2O = ?

Mass = number of mole x molar Mass

Mass of H2O = 5 x 18 = 90g

Therefore, 90g of H2O are produced

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The process by which water vapor changes to liquid water is called
andre [41]
When water is turned into vapor (a gas), that is called evaporation.

Hope this helps. :)
6 0
3 years ago
HELPP ANSWERSSS!!!!
julsineya [31]

Answer:

C.  Cs, because it has the lower ionization energy and more easily gives up its valence electrons to participate in a reaction.

Explanation:

The Cs will produce the most hydrogen gas because it has a lower ionization energy and more easily gives up its valence electrons to participate in the reaction.

  • Ionization energy is the energy required to remove the most loosely held electrons in an atom.
  • Atoms with lower ionization energy readily goes into reactions.
  • This is because they will present little barrier for the reaction to occur.
4 0
3 years ago
What is the PH of a solution whose [H+] is 0.0007
cupoosta [38]
To determine the pH of the solution. All you would have to do is use the equation:

pH = -log(H3O^+)
pH = - log(0.0007)
pH = 3.15.

Note that H+ and H3O+ are the same, it is more accurate to say H3O+ because of how the H2O molecules become after receiving the additional proton.
7 0
3 years ago
the copper anode is weighed before and after the electrolysis reation. if the copper anode is not completely dry when it is weig
Dmitrij [34]

Answer:

Faraday's constant will be smaller than it is supposed to be.

Explanation:

If the copper anode was not completely dry when its mass was measured, mass of the copper must be heavier than it should have been. Hence, the calculated Faraday’s constant would be smaller than it is supposed to be since when calculating Faraday’s Constant, the charge transferred is divided by the moles of electrons.

5 0
3 years ago
If 588 grams of FeS2 is allowed to react with 352 grams of O2 according to the following equation, how many grams of Fe2O3 are p
Rainbow [258]

<u>Answer:</u> The mass of iron (III) oxide  produced is 782.5 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For </u>FeS_2<u> :</u>

Given mass of FeS_2 = 588 g

Molar mass of FeS_2 = 120 g/mol

Putting values in equation 1, we get:

\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol

  • <u>For </u>O_2<u> :</u>

Given mass of O_2 = 352 g

Molar mass of O_2 = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol

The chemical equation for the reaction of FeS_2 and oxygen gas follows:

FeS_2+O_2\rightarrow Fe_2O_3+SO_2

By Stoichiometry of the reaction:

1 mole of FeS_2 reacts with 1 mole of oxygen gas

So, 4.9 moles of FeS_2 will react with = \frac{1}{1}\times 4.9=4.9mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, FeS_2 is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of FeS_2 produces 1 mole of iron (III) oxide

So, 4.9 moles of FeS_2 will produce = \frac{1}{1}\times 4.9=4.9moles of iron (III) oxide

Now, calculating the mass of iron (III) oxide  from equation 1, we get:

Molar mass of iron (III) oxide  = 159.7 g/mol

Moles of iron (III) oxide  = 4.9 moles

Putting values in equation 1, we get:

4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g

Hence, the mass of iron (III) oxide  produced is 782.5 grams

8 0
3 years ago
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