Question

How many grams of sulfur trioxide are produced 18 mol O2 react with sufficient sulfur? Show all your work S8 +12O 2> 8SO3

Answer 1

Answer:

$m_{SO_3}=2.31x10^4gSO_3$

Explanation:

Hello!

In this case, since the reaction between sulfur and oxygen is:

$S_8 +\frac{1}{2} O_2 \rightarrow 8SO_3$

Whereas there is a 1/2:8 mole ratio between oxygen and SO3, and we can compute the produced grams of product as shown below:

$m_{SO_3}=18molO_2*\frac{8molSO_3}{1/2molO_2} *\frac{80.06gSO_3}{1molSO_3} \\\\m_{SO_3}=23,057.3gSO_3=2.31x10^4gSO_3$

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