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gavmur [86]
3 years ago
15

an experiment is carried out to determine the formula of an oxide of mercury. the oxide is a red solid. it decomposed into its e

lements when heated at room temperature and pressure (r.t.p.). // mass of oxide of mercury: 2.16g // volume of oxygen collected at r.t.p.: 120cm³ // what is the mass of mercury produced?
Chemistry
1 answer:
Amanda [17]3 years ago
8 0

Answer:

Mass of mercury produced = 2.00 g

Explanation:

At room temperature and pressure, the temperature T = 25° C or 298.15 K while the pressure = 1 atmosphere or 760 mmHg.

Mass of oxide of mercury decomposed at room temperature and pressure = 2.16 g

Volume of oxygen produced = 120 cm³ or 0.12 dm³

One mole of any gas has a volume of 24 dm³ at room temperature and pressure.

Therefore, number of moles of oxygen produced = 0.12 dm³ / 24 dm³/mol = 0.005 moles

Mass of oxygen produced = number of moles × molar mass

Molar mass of oxygen = 32 g/mol

Mass of oxygen produced = 0.005 moles × 32 g/mol = 0.16 g

Thus, mass of mercury produced = mass of mercury oxide decomposed - mass of oxygen produced

Mass of mercury produced = 2.16 g - 0.16 g

Mass of mercury produced = 2.00 g

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How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?
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Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

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