According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-># sulfer trioxide

Answer 1

Surely it will be #0.971*mol# of #SO_3#?

We need a stoichiometic equation:

#SO_2(g) + 1/2O_2(g) rarr SO_3(g)#

The equation unequivocally tells us that the reaction of #64*g# #SO_2(g)# with #16*g# #O_2(g)# gives #80*g# #SO_3(g)#. The given masses are the molar equivalents of each gas. Again, from the stoichometric equation, sulfur was the limiting reagent (because excess dioxygen gas was specified); there was a given molar quantity of sulfur (as its dioxide), and thus a given molar quantity of the trioxide.

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Answer 2

Natalie Anton

1.457 moles of sulfur trioxide will be formed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-&gt;# sulfer trioxide

According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) #-># sulfer trioxide