Question

For the reaction below taking place at STP_S + _O2 --> _SO3 How many L of O2 would be needed to produce with 9.1 L of SO3?Give # and unit and remember sig figs.

Answer 1

1) First, let's balance the chemical equation:

_S + _O2 --> _SO3

To balance the equation, we need to equal the number of atoms of each element on the reactants and products side.

2 S + 3 O2 --> 2 SO3

Reactants side:

S - 2

O - 6

Products side:

S - 2

O - 6

2) Now to calculate, we need to know the molar volume value. This is the space occupied, in liters, by 1 mole of any matter in a gaseous state and under normal conditions of temperature and pressure (STP). The value is 22.4 liters/mol

So let's transform 9.1 L of SO3 into mole:

22.4 liters --- 1 mol

9.1 liters ---- x mol

22.4x = 9.1

x = 0.406 moles of SO3

3) Now we use the reaction proportion to know how many moles of O2 is needed:

3 moles of O2 ----- 2 moles of SO3

x moles of O2 ----- 0.40625 moles of SO3

2x = 1.21875

x = 0.609375 moles of O2

4) Now let's transform moles of O2 into liters using the molar volume value:

22.4 L ---- 1 mol

x L ----- 0.609375

x = 13.7 liters of O2

Answer: It is needed 13.7 liters of O2.