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o-na [289]
3 years ago
8

Choose the thermochemical equation that illustrates ΔH°f for Li2SO4. Choose the thermochemical equation that illustrates ΔH°f fo

r Li2SO4. 8 Li2SO4(s) → 16 Li(s) + S8(s, rhombic) + 16 O2(g) Li2SO4(aq) → 2 Li+(aq) + SO42-(aq) 2 Li+(aq) + SO42-(aq) → Li2SO4(aq) 16 Li(s) + S8(s, rhombic) + 16 O2(g) → 8 Li2SO4(s) 2 Li(s) + 1/8 S8(s, rhombic) + 2 O2(g) → Li2SO4(s)
Chemistry
1 answer:
Deffense [45]3 years ago
8 0

Answer:

2Li(s) + ⅛S₈(s, rhombic) + 2O₂(g) → Li₂SO₄(s)  

Explanation:

A thermochemical equation must show the formation of 1 mol of a substance from its elements in their most stable state,.

The only equation that meets those conditions is the last one.

A and B are wrong , because they show Li₂SO₄ as a reactant, not a product.

C is wrong because Li⁺ and SO₄²⁻ are not elements.

D is wrong because it shows the formation of 8 mol of Li₂SO₄.

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7 0
3 years ago
Still using the reaction from the video, what is the limiting reagent when 2 moles of Al are reacted
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The limiting reagent will be Al

<h3>What are limiting reagents?</h3>

They are reagents that limit the quantity of products that are formed in reactions.

From the equation of the reaction:

4Al + 3O_2 --- > 2Al_2O_3

The mole ratio of Al to O2 is 4:3.

With 2 moles of Al and 2 moles of O2, Al becomes limiting while O2 is in excess.

With 2 moles of O2, the amount of Al required should be:

                          2 x 4/3 = 2.67 moles.

With 2 moles of Al, the amount of O2 required should be:

                    2 x 3/4 = 1.5 moles

Thus, O2 is in excess by 0.5 moles.

More on limiting reagents can be found here: brainly.com/question/11848702

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