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Komok [63]
3 years ago
7

Match the effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been establishe

d in this system (N.C. = no change). 2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal pressure increases increase [O2] increase volume of chamber adding a catalyst raising the temperature
Chemistry
2 answers:
Archy [21]3 years ago
8 0

The reaction is

2SO_{2} (g)+O_{2}(g)->2SO_{3} (g)+46.8 kcal

1) the system is homogeneous with all the reactants and products are gases

2) there are three moles on reactant side and two moles on product side

i) If we increase pressure then there will be decrease in volume. It will cause increase in moles per unit volume. So if we increase pressure the equilibrium will shift towards side having less number of moles of gaseous species.

Thus system will move in forward direction.

ii) If we increase concentration of oxygen it means we are increasing concentration of reactants,this will cause system to shift in forward direction

iii) If we increase volume of chamber,  it will cause decrease in moles per unit volume, hence the system will move in the direction where number of moles of gaseous species is more, System will shift in reverse direction

iv) There is generally no effect of catalyst on equilibrium. However an increase in temperature will cause shift in reverse direction.

mestny [16]3 years ago
3 0

The effects on the concentration of SO3 gas when the following changes occur after initial equilibrium has been established in this system (N.C. = no change) by adding a catalyst.

2SO2(g) + O2(g)  2SO3(g) + 46.8 kcal

So that even though there is an addition of catalyst, no change in reactants or products has occurred because catalyst only provides a faster pathway for the reaction to occur.

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When 22.0 g NaCl and 21.0 g H, SO4 are mixed and react according to the equation below.
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Taking into account the reaction stoichiometry and limiting reagent, 26.72 grams of Na₂SO₄ are formed when 22 grams of NaOH reacts with 21 grams of H₂SO₄.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCI

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:

  • NaCl: 2 moles
  • H₂SO₄: 1 mole
  • Na₂SO₄: 1 mole
  • HCI: 2 moles

The molar mass of the compounds is:

  • NaCl: 58.45 g/mole
  • H₂SO₄: 98 g/mole
  • Na₂SO₄: 142 g/mole
  • HCI: 36.45 g/mole

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

NaCl: 2 moles ×58.45 g/mole= 116.9 grams

H₂SO₄: 1 mole ×98 g/mole= 98 grams

Na₂SO₄: 1 mole ×142 g/mole= 142 grams

HCI: 2 moles ×36.45 g/mole= 72.9 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this reaction</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 116.2 grams of NaCl reacts with 98 grams of H₂SO₄, 22 grams of NaCl reacts with how much moles of H₂SO₄?

mass of H_{2} SO_{4} =\frac{22 grams of NaClx98 grams of H_{2} SO_{4} }{116.2 grams of NaCl}

<u><em>mass of H₂SO₄= 18.55 grams</em></u>

But 18.55 grams of H₂SO₄ are not available, 21 grams are available. Since you have less moles than you need to react with 22 grams of NaCl, H₂SO₄ will be the limiting reagent.

<h3>Mass of Na₂SO₄ produced</h3>

The following rules of three can be applied, considering the limiting reagent: if by reaction stoichiometry 116.9 grams of NaCl form 142 grams of Na₂SO₄, 22 grams of NaCl form how much mass of Na₂SO₄?

mass of Na_{2}S O_{4} =\frac{22 grams of NaClx142 grams of Na_{2}S O_{4}}{116.9 grams of NaCl}

<u><em>mass of Na₂SO₄= 26.72 grams</em></u>

Then, 26.72 grams of Na₂SO₄ are formed when 22 grams of NaOH reacts with 21 grams of H₂SO₄. The correct answer is first option.

Learn more about the reaction stoichiometry:

<u>brainly.com/question/24741074</u>

<u>brainly.com/question/24653699</u>

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Answer:

mass of H₂O = 54.06 g

Explanation:

balanced chemical equation

    Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(l)

given data

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 mas of H₂O = ?

 Molar mass of H₂O = 18.02 g/mol

we find the mass of H₂O by:

    mass of H₂O = moles × molar mass

    mass of H₂O = 3 mol × 18.02 g/mol

    mass of H₂O = 54.06 g

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