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2 years ago

How many molecules of O₂ will be required to produce 28.8 g of water? 2H₂ + O₂ ---> 2H₂O

Chemistry

1 answer:

juin [17]2 years ago

4 0

To produce 28.8 g of water 4.81 x 10²³ molecules of O₂ will be required

<h3>What is limiting reagent ?</h3>

The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

According to given equation;

2H₂ + O₂ ---> 2H₂O

If 36 g of H₂O (2 mole) is produced by 1 mole O₂ i.e, 32 gram.

Therefore, 28.8 g of H₂O is produced by ;

32/36 x 28.8 = 25.6 g of O₂

Number of molecules of O₂ = given weight / molecular weight x 6.023 x 10²³

= 25.6 / 32 x 6.023 x 10²³

= 4.81 x 10²³ molecules of O₂

Therefore,4.81 x 10²³ molecules of O₂ will be required to produce 28.8 g of water.

Learn more about Limiting reagent here ;

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Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

Which is NOT a way that Carbon is released in to the atmosphere?

amid [387]

Answer:

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Explanation:

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2 years ago

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In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga

Reptile [31]

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

$N_2+3H_2\rightarrow 2NH_3$ Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol$

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

$\frac{2}{1}\times 12.1597 mol=24.3194 mol$ of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

6 0

3 years ago

Need structural formula for each 12, HOCH a CH 1 Dipherylamine.

guajiro [1.7K]

12) Ethylene glycol and <span>Diphenylamine
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<span>attached images
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5 0

3 years ago

A chemist describes a particular experiment in this way: "0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol o

AleksandrR [38]

Description:

<span>"0.0400 mol of H2O2 decomposed into 0.0400 mol of H2O and 0.0200 mol of O2."

This means that a certain amount of H2O2 (0.0400 mol) decomposed or was broken down into two components, 0.04 mol of H2O and 0.02 mol of O2. To examine the system, we need a balanced equation:

H2O2 ---> H2O + 0.5O2

The final concentrations of the system indicates that the system is in equilibrium. </span>

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3 years ago