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MatroZZZ [7]
2 years ago
5

Calcium carbonate decomposes into calcium oxide and carbon dioxide when heated. Calculate the mass of carbon dioxide released wh

en 300 g of calcium carbonate is heated.
Chemistry
1 answer:
erastovalidia [21]2 years ago
6 0
<h2>Answer: 131.9 g</h2>

<h3>Explanation:</h3>

<u>Write a Balanced Equation for the decomposition</u>

CaCO₃     →     CaO    +    CO₂

<u></u>

<u>Find Moles of CO₂ Produced</u>

Since the mole ratio of  CaCO₃  to CO₂ is 1 to 1,

the moles of CaCO₃ = moles of CO₂

moles of CaCO₃   = mass ÷ molar mass

                             = 300 g ÷ 100.087 g/mol

                             = 2.997 moles

∴ moles of CO₂ = 2.997 moles

<u>Determine Mass of CO₂</u>

Mass = moles × molar mass

         = 2.997 mol × 44.01 g/mol

         = 131.9 g

<u></u>

<h3>∴ when 300 g of calcium carbonate is decomposed, it produces 131.9 g of carbon dioxide.</h3>
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Decreases

Explanation:

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Calculate the number of C, H, and O atoms in 1.50 g of glucose, a sugar
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Chemical formula of the glucose: C₆H₁₂O₆

We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u

atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol


2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g----------------------  x

x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles

we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)

3)We calculate the number of molecules:

Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles--------        x

x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.

4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²

number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .

number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²

Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
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How many moles of NH3 would be formed from the complete reaction of 16.0 g H2?
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Taking into account the reaction stoichiometry, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

N₂ + 3 H₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • N₂: 1 mole
  • H₂: 3 moles
  • NH₃: 2 moles

The molar mass of the compounds is:

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  • H₂: 2 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • N₂: 1 mole ×14 g/mole= 14 grams
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<h3>Mass of NH₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 2 moles of NH₃, 16 grams of H₂ form how many moles of NH₃?

moles of NH_{3}= \frac{16 grams of H_{2} x2moles of NH_{3}}{6 grams of H_{2}}

<u><em>moles of NH₃= 5.33 moles</em></u>

Then, 5.33 moles of NH₃ are formed from the complete reaction of 16 grams of H₂.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

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