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jekas [21]
2 years ago
9

How would I Determine the number of moles in 3.51 x 10^23 formula units of CaCl2

Chemistry
1 answer:
Andrej [43]2 years ago
8 0

Answer:

by using this formula you will get it

Explanation:

number of mole = number of particles÷ Avogadro's number

n=3.51×10^23÷ 6.02×10^23

n = 0.58 moles

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Using the following equation, 2C2H6 +7O2 --&gt;4CO2 +6H2O, if 2.5g C2H6 react with 170g of O2, how many grams of water will be p
kirill [66]

The mass of water (H₂O) that would be produced is 4.5 g

<h3>Stoichiometry </h3>

From the question, we are to determine the mass of water that would be produced.

From the given balanced chemical equation

2C₂H₆ +7O₂ → 4CO₂ +6H₂O

This means

2 moles of C₂H₆ reacts with 7 moles of O₂ to produce 4 moles of CO₂ and 6 moles of H₂O

Now, we will determine the number of moles of each reactant present

  • For Ethane (C₂H₆)

Mass = 2.5 g

Molar mass = 30.07 g

Using the formula,

Number\ of\ moles = \frac{Mass}{Molar\ mass}

Number of moles of C₂H₆ present = \frac{2.5}{30.07}

Number of moles of C₂H₆ present = 0.08314 mole

  • For Oxygen (O₂)

Mass = 170g

Molar mass = 31.999 g/mol

Number of moles of O₂ present = \frac{170}{31.999}

Number of moles of O₂ present = 5.3127 moles

Since

2 moles of C₂H₆ reacts with 7 moles of O₂

Then,

0.08314 mole of C₂H₆ will react with \frac{7 \times 0.08314 }{2}

 \frac{7 \times 0.08314 }{2} = 0.58198 mole

Therefore,

0.08314 mole of C₂H₆ reacts with 0.58198 mole of O₂ to produce 3 × 0.08314 moles of H₂O

3 × 0.08314 = 0.24942 mole

Thus, the number of moles of water (H₂O) produced is 0.24942 mole

Now, for the mass of water that would be produced,

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of water = 18.015 g/mol

Then,

Mass of water that would be produced = 0.24942 × 18.015

Mass of water that would be produced = 4.4933 g

Mass of water that would be produced ≅ 4.5 g

Hence, the mass of water (H₂O) that would be produced is 4.5 g

Learn more on Stoichiometry here: brainly.com/question/14271082

3 0
2 years ago
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