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

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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