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earnstyle [38]
3 years ago
6

How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl3?

Chemistry
1 answer:
lana66690 [7]3 years ago
3 0

Answer:

6.142 moles of NaCl

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2AlCl3 + 3Na2S —> Al2S3 + 6NaCl

Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:

Mass mass of Na2S = 78.048g/mol

Mass of Na2S = 239.7g

Number of mole Na2S =..?

Mole = Mass /Molar Mass

Number of mole Na2S = 239.7/78.048 = 3.071 moles

Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:

From the balanced equation above,

3 moles of Na2S reacted to produce 6 moles of NaCl.

Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl

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An open flask sitting in a lab refrigerator looks empty, but it is actually filled with a mixture of gases called air. If the fl
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9.39 × 10²² molecules

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P . V = n . R . T

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n=\frac{P.V}{R.T} =\frac{1atm.3.50L}{(0.08206atm.L/mol.K).273.15K} =0.156mol

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THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.

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An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.

The kinetic-molecular theory as it applies to gases has five basic assumptions.

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