Answer:
Option a. Oxygen gas is limiting
Explanation:
First, let us write a balanced equation for the reaction. This is illustrated below:
2H2 + O2 —> 2H2O
Now, we were told from the question that equal masses of H2 and O2 reacted.
Let the mass of H2 and O2 be 2g each.
Next let us covert this mass to mole as shown below:
Molar Mass of H2 =2x1 = 2g/mol
Assumed mass of H2 = 2g
Number of mole =?
Number of mole = Mass /Molar Mass
Number of mole of H2 = 2/2 = 1mole
Molar Mass of O2 = 2x16 = 32g/mol
Assumed mass of O2 = 2g
Number of mole =?
Number of mole = Mass /Molar Mass
Number of mole of O2 = 2/32 = 0.063mole
Data obtained from our calculations:
Number of mole H2 = 1mole
Number of mole of O2 = 0.063mole
Now to obtain the limiting reactant, do the following:
From the equation,
2moles of H2 required 1mole of O2.
Therefore, 1mole of H2 will require = 1/2 = 0.5mole of 02.
This amount (0.5mole) of O2 obtained is far greater than the amount (i.e 0.063mole) of O2 earlier calculated for. Therefore it is not acceptable.
Now let us turn the tide around.
From the equation,
2moles of H2 required 1mole of O2.
Therefore, Xmol of H2 will require 0.063mole of O2 i.e
Xmol of H2 = 2 x 0.063 = 0.126mol
This amount (0.126mole) of H2 obtained is far lesser than the amount (i.e 1 mole) of H2 earlier calculated for. Therefore it is acceptable. This implies that H2 is the excess reactant and O2 is the limiting reactant.
