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ollegr [7]
3 years ago
15

When 1.50 mol of solid aluminum metal reacts with 3.00 mol of chlorine gas (CI2), aluminum chloride solid is formed. Which is th

e limiting reactant?
Chemistry
1 answer:
expeople1 [14]3 years ago
7 0

Answer:

Aluminum is limiting reactant

Explanation:

Data Given:

Amount aluminum metal (Al) = 1.5 mole

chlorine gas (Cl₂) = 3 mole

limiting reactant = ?

Solution:

limiting reactant: The amount of reactant that consume first in a reaction is termed as limiting reagent or limiting reactant.

Excess reagent: the amount of reactant that remain unreacted at the end of the reaction is termed as excess reagent.

Now to look for the limiting regent we have to look at reaction

Reaction:

           2 Al + 3 Cl₂  --------> 2 AlCl₃

from this balance reaction we come to know that 2 mole of Aluminum metal react with 3 mole of chlorine gas. It means that amount of aluminum is less and it will consume first and act as limiting reactant.

To know with how much chlorine gas it will react apply unity formula

                2 mole of Al ≅ 3 mole of Cl₂

               1.5 mole of Al ≅ X mole of Cl₂

So,

                mole of Cl₂ =   3 mole x  1.5 mole / 2 mol

                mole of Cl₂ =   2.25 mol

So, 1.5 mol Aluminum consumed fully and react with 2.25 mol of Chlorine and 0.75 mol remain unreacted.

Aluminum is limiting reactant.

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3 years ago
Explain why different liquids boil at the different temperatures
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The boiling point of the fluid depends on the intermolecular forces between the fluid atoms and molécules, as these forces must be disrupted to switch from a fluid to a gas. The stronger the intermolecular forces, the greater the point of boiling.

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3 years ago
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Polymerization is the process of linking smaller molecules to form long chains of higher molecular weight. True or False
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True is correct answer.


Polymerization is the process of linking it has a smaller molecules to form has a long chains of higher molecular weight.

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3 years ago
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How many moles of hydrogen gas will be produced if 5.00 moles of zinc reactions with an excess amount of sulfuric acid
harkovskaia [24]

Answer:

Moles of Hydrogen produced is 5 moles

Explanation:

The balanced Chemical equation for reaction between zinc and sulfuric acid is :

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

This equation tells that ; when  1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

1\ mole\ zinc\rightarrow 1\ mole\ H_{2}

So,

5\ mole\ zinc\rightarrow 5\ mole\ H_{2}

Hence moles of Hydrogen produced is 5 moles

3 0
3 years ago
Use standard enthalpies of formation to calculate Δ H ∘ rxn for each reaction. MISSED THIS? Read Section 7.9; Watch KCV 7.9, IWE
Eva8 [605]

Answer:

Standard Heat of Reaction 1 = -136.2 kJ/mol

Standard Heat of Reaction 2 = -41.166 kJ/mol

Standard Heat of Reaction 3 = -136.07 kJ/mol

Standard Heat of Reaction 4 = 279.448kJ/mol

Explanation:

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

The required standard heat of formation for each of the reactants and product above, as obtained from literature is listed below.

C₂H₄ (g), 52.5 kJ/mol

H₂ (g), 0 kJ/mol

C₂H₆ (g), -83.7 kJ/mol

CO (g), -110.525 kJ/mol

H₂O (g), -241.818 kJ/mol

H₂ (g), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

NO₂ (g), 33.2 kJ/mol

H₂O (l), -285.8 kJ/mol

HNO₃ (aq), -206.28 kJ/mol

NO (g), 90.29 kJ/mol

Cr₂O₃ (s), -1128.4 kJ/mol

CO (g), -110.525 kJ/mol

Cr (s), 0 kJ/mol

CO₂ (g), -393.509 kJ/mol

Note that

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

C₂H₄ (g) + H₂ (g) → C₂H₆ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×-83.7) = -83.7 kJ/mol

ΔH∘(reactants) = (1×52.5) + (1×0) = 52.5 kJ/mol

ΔH∘(rxn) = -83.7 - 52.5 = -136.2 kJ/mol

CO (g) + H₂O (g) → H₂ (g) + CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (1×0) + (1×-393.509) = -393.509 kJ/mol

ΔH∘(reactants) = (1×-110.525) + (1×-241.818) = -352.343 kJ/mol

ΔH∘(rxn) = -393.509 - (-352.343) = -41.166 kJ/mol

3NO₂ (g) + H₂O (l) → 2HNO₃ (aq) + NO (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×-206.28) + (1×90.29) = -322.27 kJ/mol

ΔH∘(reactants) = (3×33.2) + (1×-285.8) = -186.2 kJ/mol

ΔH∘(rxn) = -322.27 - (-186.2) = -136.07 kJ/mol

Cr₂O₃ (s) + 3CO (g) → 2Cr (s) + 3CO₂ (g)

ΔH∘(rxn) = ΔH∘(products) - ΔH∘(reactants)

ΔH∘(products) = (2×0) + (3×-393.509) = -1,180.527 kJ/mol

ΔH∘(reactants) = (1×-1128.4) + (3×-110.525) = -1,459.975 kJ/mol

ΔH∘(rxn) = -1,180.527 - (-1,459.975) = 279.448 kJ/mol

Hope this Helps!!!

4 0
3 years ago
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