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tresset_1 [31]
3 years ago
5

In the reaction below, if 5.71 g of sulfur is reacted with 10.0 g of oxygen, how many grams of sulfur trioxide will be produced?

Chemistry
1 answer:
yawa3891 [41]3 years ago
3 0

Answer:

14.3 g SO₃

Explanation:

2S + 3O₂ → 2SO₃

First, find the limiting reactant.  To do that, calculate the mass of oxygen needed to react with all the sulfur.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂

0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂

There are 10.0 g of O₂, so there's enough oxygen.  The limiting reactant is therefore sulfur.

Use the mass of sulfur to calculate the mass of sulfur trioxide.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃

0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃

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2. The number of mole of Chlorine pentafluoride, ClF5, is 0.3mole

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Now substituting the value of n in equation 1, we have:

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Now we can obtain the mass of Chlorine pentafluoride ClF5 as follow:

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R (gas constant) = 0.082atm.L/Kmol

m (mass of Chlorine pentafluoride) =?

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Cross multiply to express in linear form as shown below:

m x 0.082 x 256 = 0.180x35x130.5

Divide both side by 0.082 x 256

m = (0.180x35x130.5) /(0.082x256)

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Mass of ClF5 = 39.16g

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Number of mole = Mass /Molar Mass

Mole of ClF5 = 39.16/130.5g

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