Question

A flask contains a gaseous mixture of 3.01mol of hydrogen and 5.84mol of oxygen gases. When these gases react, steam is produced.

a) Balance the equation for this reaction. H2(g) + O2(g) H2O(g)

b) What is the limiting reagent?
c) How many grams of steam are produced?
d) How many moles of limiting reagent remain unreacted? e) How many moles of excess reagent remain unreacted?

Answer 1

The reaction equation is expressed as

H2(g) + O2(g) H2O(g)

Hence the balanced equation for this reaction is

2H2(g) + O2(g) 2H₂O(g)

In this reaction 1 mole of O2(g) reacts with 2 moles of H2(g).

But you have

3.01 mol of hydrogen and 5.84 mol of oxygen gases.

So, 5.84 mol of oxygen gases will react with (2 x 5.84) mol of hydrogen gases.

  5.84 mol of oxygen gases will react with 11.68 mol of hydrogen gases.

Or

2 moles of H2(g) reacts with 1 mole of O2(g).

  1 moles of H2(g) reacts with 1/2 mole of O2(g).

3.01 moles of H2(g) reacts with 3.01/2 mole of O2(g).

3.01 moles of H2(g) reacts with 1.505 mole of O2(g).

Hence, O2(g) is used in excess. So, limiting reagent is H2(g).

Again,

The balanced equation for this reaction is

2H2(g) + O2(g) 2H₂O(g)

In this reaction 2 moles of H2(g) produces 2 moles of H₂O(g).

1 moles of H2(g) produces 1 moles of H₂O(g).

3.01 moles of H2(g) produces 3.01 moles of H₂O(g).

Limiting reagents are used completely, so no moles of limiting reagent remain unreacted.

Excess reagent left = Total moles used - Moles reacted

= 5.840 mol - 1.505 mole

= 4.335 mole