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

Based on the standards of units conversion, to convert from micrometer to meter, we multiply by 10^-6.
Since we there is a square (10^2) to consider, then to convert from micrometer squared to meter squared, we will multiply by (10^-6)^2 as follows:
1.5 μm2 = 1.5 x (10^-6)^2 = 1.5 x 10^-12 meter sqaures