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Slav-nsk [51]
3 years ago
10

What is the theoretical yield of Ca(OH)2, in grams, if 31.8 g of CaO is hydrolyzed (reacted) in an excess of water?

Chemistry
1 answer:
Fiesta28 [93]3 years ago
6 0

The theoretical yield of Ca(OH)₂ : 42.032 g

<h3>Further explanation</h3>

Given

31.8 g of CaO

Required

The theoretical yield of Ca(OH)₂

Solution

Reaction

CaO + H₂O⇒Ca(OH)₂

mol CaO (MW=56 g/mol) :

= mass : MW

= 31.8 g : 56 g/mol

= 0.568

From equation, mol Ca(OH)₂ = mol CaO = 0.568

Mass Ca(OH)₂ (MW=74 g/mol) :

= 0.568 x 74

= 42.032 g

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A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8
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16.06 L was the initial volume of the balloon.

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Using Avogadro's law:

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Answer:

2.067 L ≅ 2.07 L.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>

It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.

  • At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u><em>using cross multiplication:</em></u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 3.1 L.

∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.

  • To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:

<u><em>Using cross multiplication:</em></u>

3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.

0.1384 mol of O₂ produce → ??? mol of SO₂.

∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.

  • Again, using cross multiplication:

1.0 mol of SO₂ represents → 22.4 L, at STP.

0.09227 mol of SO₂ represents → ??? L.

∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.

8 0
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