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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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7 0
3 years ago
Read 2 more answers
A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out
scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<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 lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol

  • The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = \frac{2}{1}\times 0.0021=0.0042mol of potassium bromide

  • Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g

  • To calculate the percentage composition of KBr in the mixture, we use the equation:

\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%

Hence, the percent by mass of KBr in the mixture is 9.996 %

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3 years ago
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boyakko [2]

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4 0
3 years ago
Name each subatomic particle, its charge, and its location in an atom.
Tpy6a [65]
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4 0
3 years ago
What is the percent composition by mass of oxygen in Ca(NO3)2 (gram-formula= 164 g/mol)?
Neko [114]
To find this, we will use this formula:

Molar mass of element
------------------------------------ x 100
Molar mass of compound

So, first lets calculate the mass of the compound as a whole. We use the atomic masses on the periodic table to determine this.

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When we add all of those numbers up together, we get 164.086. That is the molar mass for the whole compound. However, we are trying to figure out what percent of the compound oxygen makes up. From the molar mass, we know that 95.994 of the 164.086 is oxygen. Lets plug those numbers into our equation!

95.994
-----------
164.086

When we divide those two numbers, we get .585. When we multiply that by 100, we get 58.5.

So, the percent compostition of oxygen in Ca(NO3)2, or, calcium nitrate, is 58.5%.
5 0
3 years ago
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