At STP, how many liters of oxygen molecules would be required to burn 5.5 g of C 6 H 6 ?

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For this question, we first need to write the balanced chemical equation for the reaction between benzene (C6H6) and oxygen.

The reaction is

C6H6 + O2 -> CO2 + H2O + heat.

The balanced chemical reaction for this interaction between benzene and oxygen can be written as

C6H6 + 15/2...

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For this question, we first need to write the balanced chemical equation for the reaction between benzene (C6H6) and oxygen.

The reaction is

C6H6 + O2 -> CO2 + H2O + heat.

The balanced chemical reaction for this interaction between benzene and oxygen can be written as

C6H6 + 15/2 O2 -> 6CO2 + 3H2O + heat.

Here, 1 mole of benzene reacts with 15/2, or 7.5, moles of oxygen to produce 6 moles of carbon dioxide and 3 moles of water, along with heat.

In the given question, we are provided with 5.5 g of benzene. How many moles of benzene are there in 5.5 g?

For this, we need to first determine the molecular weight of benzene. Carbon has an atomic mass of 12 g/moles, and hydrogen has an atomic mass of 1 g/moles.

Since, benzene is C6H6, its molecular weight can be determined as

6 x atomic mass of carbon + 6 x atomic mass of hydrogen = 6 x 12 + 6 x 1 = 78 g/moles.

Thus, 78 g of benzene is equivalent to 1 mole,

and 5.5 g of benzene is equivalent to 1/78 x 5.5 moles = 0.071 moles.

Since 1 mole of benzene reacts with 7.5 moles of oxygen, 0.071 moles of benzene will react with 7.5 x 0.071 moles = 0.53 moles of oxygen.

At STP (standard temperature and pressure), 1 mole of an ideal gas occupies 22.4 l of volume.

Thus, 0.53 moles of oxygen will occupy 22.4 x 0.53 = 11.87 l of volume.

Hope this helps.

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