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Kipish [7]
3 years ago
11

how many grams of calcium oxide will be produced in a closed vessel containing 20.0 kg of calcium and 20.0 kg of oxygen gas if t

he reaction goes to completion?
Chemistry
1 answer:
sergejj [24]3 years ago
8 0

Answer:

27984.1311 g  

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

<u>For calcium :- </u>

Mass of calcium = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of calcium = 40.078 g/mol

Thus,

Moles= \frac{20000\ g}{40.078\ g/mol}

Moles\ of\ calcium= 499.0269\ mol

<u>For oxygen gas :- </u>

Mass of oxygen gas = 20.0 kg = 20000 g ( 1 kg = 1000 g )

Molar mass of oxygen gas = 31.999 g/mol

Thus,

Moles= \frac{20000\ g}{31.999\ g/mol}

Moles\ of\ oxygen\ gas= 625.0195\ mol

According to the given reaction:

2Ca+O_2\rightarrow 2CaO

2 moles of calcium reacts with 1 moles of oxygen gas

Also,

1 mole of calcium react with 1/2 mole of oxygen gas

So,

499.0269 mole of calcium react with \frac{1}{2}\times 499.0269 mole of oxygen gas

Moles of oxygen gas = 249.5135 moles

Available moles of oxygen gas = 625.0195 moles

Limiting reagent is the one which is present in small amount. Thus, calcium is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

2 moles of calcium produces 2 moles of calcium oxide

So,

1 mole of calcium produces 1 mole of calcium oxide

Also,

499.0269 mole of calcium produces 499.0269 mole of calcium oxide

Moles of calcium oxide = 499.0269 moles

Molar mass of calcium oxide = 56.0774 g/mol

Mass of calcium oxide = Moles × Molar mass = 499.0269 × 56.0774 g = 27984.1311 g  

<u>27984.1311 g  of calcium oxide will be produced.</u>

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Answer: _{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

General representation of an element is given as:   _Z^A\textrm{X}

where,

Z represents Atomic number

A represents Mass number

X represents the symbol of an element

General representation of alpha decay :

_Z^A\textrm{Sr}\rightarrow _{Z-2}^{A-4}Y+_2^4\alpha

The  balanced nuclear equation when the isotope Strontium-90 decays by Q-  decay is :

_{38}^{90}\textrm{Sr}\rightarrow _{36}^{86}\textrm{Kr}+_2^4\textrm{He}

4 0
3 years ago
A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o
Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

c=\frac{n}{V}

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol

Now, the total volume of this solution can be found by adding the volume values of each component:

V_total=46.2 mL+80.5 mL=126.7 mL

Finally, dividing the moles found by the total volume will yield the final molarity:

c_{final}=\frac{138.62 mmol}{126.7 mL}= 1.09 M

6 0
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How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate
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5 0
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Calculate the number of moles in the following: 2.8 X 10^24 atoms of Cl2
vova2212 [387]

Answer:

<h3>The answer is 4.65 moles</h3>

Explanation:

To find the number of moles given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question

N = 2.8 × 10²⁴ atoms of Cl2

So we have

n =  \frac{2.8 \times  {10}^{24} }{6.02 \times  {10}^{23}  }  \\  = 4.65116279069...

We have the final answer as

<h3>4.65 moles</h3>

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