if 0.40 mol of h2 and .15 mol of o2 were to reat as completely as possible to produce h20, what mass of the reactant would remain

0.2g

Explanation:

Given parameters:

Number of moles of H₂  = 0.4mol

Number of moles of O₂  = 0.15mol

Unknown:

Mass of reactant that would remain = ?

Solution:

To solve this problem, we need to know the limiting reactant which is the one in short supply in the given reaction.

  The expression of the reaction is :

                2H₂  + O₂  →   2H₂O

                    2 mole of H₂ will combine with 1 mole of O₂

But given;    0.4 mole of H₂ we will require  = 0.2mole of O₂

The given number of oxygen gas is 0.15mole and it is the limiting reactant.

Hydrogen gas is in excess;

       1 mole of oxygen gas will combine with 2 mole of hydrogen gas

    0.15 mole of oxygen gas will require 0.15 x 2  = 0.3mole of hydrogen gas

Now, the excess mole of hydrogen gas  = 0.4 mole  - 0.3 mole  = 0.1mole

  Mass of hydrogen gas  = number of mole x molar mass

  Molar mass of hydrogen gas  = 2(1) = 2g/mol

   Mass of hydrogen gas  = 0.1 x 2 = 0.2g

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.