In the rusting reaction, 2.4 mol of iron reacts with 1.8 mol of oxygen, as determined by the 4:3 mole ratio between iron and oxygen in the balanced equation.
To determine the number of moles of oxygen that react with 2.4 mol of iron in the rusting reaction:
1. Examine the balanced chemical equation:
4Fe(s) + 3O2(g) → 2Fe2O3(s)
2. Identify the mole ratio between iron and oxygen:
From the balanced equation, we can see that the mole ratio of iron to oxygen is 4:3. This means that for every 4 moles of iron, 3 moles of oxygen are required.
3. Use the mole ratio to calculate the number of moles of oxygen:
Given that we have 2.4 mol of iron, we can set up a proportion using the mole ratio:
4 mol Fe / 3 mol O2 = 2.4 mol Fe / x mol O2
Cross-multiplying, we get:
4 mol Fe * x mol O2 = 2.4 mol Fe * 3 mol O2
Simplifying the equation:
4x mol O2 = 7.2 mol O2
4. Solve for x to find the number of moles of oxygen:
Dividing both sides of the equation by 4:
x mol O2 = 7.2 mol O2 / 4
The number of moles of oxygen that react with 2.4 mol of iron is 1.8 mol.
The question probable may be:
When iron rusts in air, iron(III) oxide is produced. How many moles of oxygen react with 2.4 mol of iron in the rusting reaction?
4Fe( s ) + 3O 2 ( g ) → 2Fe 2 O 3 ( s ).