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3 years ago

Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. S+(sub2) -----> __SO(sub3)

1 answer:

6 0

2S + 3O2 ==> 2SO3
Let's start off with the unbalanced equation and count atoms of each element on both sides of the arrow.
S + O2 ==> SO3
There's 1 sulfur atom on both sides which is good. However, there's 2 oxygen atoms on the left and 3 on the right. So calculate the Least common multiple (LCM) of 2 and 3 which is 6. And adjust the coefficients on both sides to have 6 oxygen atoms on each side, giving:
S + 3O2 ==> 2SO3
Now we have 6 oxygen atoms on each side which is good, but there's only 1 sulfur atom on the left and 2 on the right. Since the LCM of 1 and 2 is 2, all we need to do is adjust the coefficient on the left to get 2 sulfur atoms there. So:
2S + 3O2 ==> 2SO3
We now have 2 sulfur atoms on each side of the equation and 6 oxygen atoms on each side, so the resulting equation is balanced.

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The mass of oxygen reacted/required in this reaction is obtained as 48g.

<h3>What is stoichiometry?</h3>

The term stoichiometry has to do with mass- volume or mass - mole relationship which ultimately depends on the balanced reaction equation.

Now, we have the reaction; S + O2 ------>SO2

If 1 mole of sulfur dioxide contains 22.4 L

x moles of sulfur dioxide contains 33.6L

x = 1.5 moles of sulfur dioxide.

Since the reaction is 1:1, the number if moles of oxygen required/reacted is 1.5 moles.

Mass of oxygen required/reacted = 1.5 moles * 32 g/mol = 48g

Learn more anout stoichiometry: brainly.com/question/9743981

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$Molarity = \frac{no. of moles}{Volume (in L)}$

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Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. __S+__(sub2) -----> __SO(sub3)

Sulfur reacts with oxygen to produce sulfur trioxide gas. Balance the equation. S+(sub2) -----> __SO(sub3)