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

How many moles of NaCl are produced if 239.7 grams of Na2S reacts with plenty of AlCl-example-1

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Answer:

6.142 moles of NaCl

Step-by-step 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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