Answer:
1) 1.807 * 10^24 molecules of O2
2) 6 oxygen atoms required
3) There are 2 moles of Al2O3 formed
4) Mole ratio Al to oxygen = 2/3 = 0.6667
Explanation:
Step 1: The balanced equation
4Al(s) + 302(g) → 2Al203 (s)
We see that we need to consume 4 moles of Aluminium and 3 moles of oxygen to produce 2 moles of Al2O3
<em>How many molecules of O2 are used in the reaction?</em>
we have 3 moles of 02
The number of molecules of 02 in the reaction = 3 moles * 6.022 *10^23 = 1.807 * 10^24 molecules of O2
<em>How many oxygen atoms are required?</em>
Since we have 3 moles of 02 this means there is 3*2 = 6 oxygen atoms required
<em>How many moles of Al2O3 are formed?</em>
There are 2 moles of Al2O3 formed
<em>What is the mole ratio of Al to O?</em>
In the reactant side 4 Al reacts with six oxygen atoms to for two molecules of Al2O3
Mole ratio Al to oxygen = (No of atoms of Al in Al2O3)/ (Number of atoms of oxygen in Al2O3)
Mole ratio Al to oxygen = 2/3 = 0.6667
