Equal masses (in grams) of hydrogen gas and oxygen gas are reacted to form water. Which substance is limiting?

a. Oxygen gas is limiting

b. Hydrogen gas is limiting.

c. Water is limiting.

d. Nothing is limiting.

e. More information is needed to answer this question

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.

a. Oxygen gas is limiting

Explanation:

hydrogen gas and oxygen gas are reacted to form water

2H₂ + O₂ → 2H₂O

the above balanced equation shows that 2 moles of H₂ is required for 1 mole of O₂

Given equal masses of H₂ and O₂

assuming 'x' gm for each, no. of moles of each gas =

no. of moles of H₂ = x/2 = 0.5x moles

no. of moles of O₂ = x/32 = 0.031x moles

This shows that no. of moles of O₂ is very less so O₂ will become the limiting reagent.