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valkas [14]
3 years ago
5

5. Sodium peroxide reacts vigorously with water in the following unbalanced equation:

Chemistry
1 answer:
Rasek [7]3 years ago
7 0

Answer:

30.7 g of O₂ are produced

Explanation:

This is a redox reaction, where the peroxide is reduced to oxygen and oxidized to hydroxide. The equation is this one:

2Na₂O₂ + 2H₂O → 4NaOH + 3O₂

We assume that water is in excess, so the peroxide is the limiting reagent.

We convert the mass to moles: 50 g . 1mol / 78 g = 0.641 moles

2 moles of peroxide can produce 3 moles of oxygen (according to stoichiometry)

Then, 0.641 moles of peroxide will produce (0.641 . 3 ) /2 = 0.961 moles O₂

Finally, we convert the moles to mass: 0.961mol . 32 g / 1mol = 30.7 g of O₂

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Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O

Cr goes from +6 to +3 by gaining 3 e

Hg goes from 0 to +2 by losing 2 e

we need 3 Hg for every 2 Cr

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Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

2 Cr on the right and left

Net 12 positive charges on the right and the left

3 Hg on the right and left

14 H on the right and left

the equation is balanced

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