Question

When 134.4 L of H2 combines with 89.6 L of O2 at STP then the number of moles of H2O
formed is ?​
a.3 mole
b. 6 mole
c. 2.5 mole
d.2 mole ​

Answer 1

Answer:

6 moles

Explanation:

In this case H2 is the limiting reagent.

let's take the number of moles produced as n.

2 x 22.4L of H2 = 2 moles of H2O

134.4 L of H2. = n moles of H2O

n = 2 x 134.4/2 x 22.4

n = 6 moles.

Answer 2

Answer:

(b) 6mole

Explanation:

at STP,

no.of mol.of H2 = 134.4L/22.4L/mol = 6mol

no.of mol of O2 = 89.6L/22.4L/mol = 4 mol

2H2 + O2 = 2H2O

from stoichiometry of balanced reaction

2mol of H2 = 1 mol of O2

6 mol of H2 = 3 mol of O2

so, H2 is the limiting reactant.

Now, 2 mol of H2 produces 2 mol of H2O

then , 6mole of H2 produces 6mol of H2O