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Ivanshal [37]
3 years ago
11

How many sulfur atoms are there in 21.0 g of al2s3?

Chemistry
1 answer:
ValentinkaMS [17]3 years ago
3 0

Given the mass of aluminum sulfideAl_{2}S_{3} = 21.0 g

Molar mass of Al_{2}S_{3} = 2 * Molar mass of Al + 3 *molar mass of S = 2(27g/mol)+3(32.g/mol)=150.g/mol

Calculating the moles of Al_{2}S_{3}:

21.0 g *\frac{1 mol}{150 g} = 0.14 mol Al_{2}S_{3}

Each mole Al_{2}S_{3} has 3 mol S.

Each mole S constitutes 6.022*10^{23} atoms of S

Calculating atoms of S in 0.14 mol Al_{2}S_{3}:

0.14 mol Al_{2}S_{3} *\frac{3mol S}{1 mol Al_{2}S_{3}  }*\frac{6.022*10^{23}atoms S }{1 mol S}

= 2.53 * 10^{23} atoms of S

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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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0.088 mole of Al.

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

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Mole of AlBr₃ =.?

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Mole of AlBr₃ = 23.6 / 267

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

This is given below:

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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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