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3 years ago

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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Read 2 more answers

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Let's check Electronic configuration of N in ground state

$\\ \sf\longmapsto 1s^22s^22p^3$

In exited state

$\\ \sf\longmapsto 1s^22s^22px^12py^12pz^2$

3 p orbitals

In hydrogen case

$\\ \sf\longmapsto 1s^1$

1 s orbital

Hence Hybridization

$\\ \sf\longmapsto s-p-p-p$

$\\ \sf\longmapsto sp^3$

Structure is tetrahedral .

Spare way:-

Bond pairs==3=>Bonding electrons=6
Lone pair=1=>Anti bonding electrons=2

Hybridization

$\\ \sf\longmapsto \dfrac{1}{2}[\sigma+\sigma *]$

$\\ \sf\longmapsto \dfrac{1}{2}[6+2]$

$\\ \sf\longmapsto \dfrac{1}{2}[8]$

$\\ \sf\longmapsto 4$

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