Answer the questions about the following reaction.
4al(s) + 302(g) → 2a1203 (s)
how many molecules of o2 are used in the reaction?
how many oxygen atoms are required?
how many moles of al2o3 are formed?
what is the mole ratio of al to o?

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

How many molecules of O2 are used in the reaction?

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

How many oxygen atoms are required?

Since we have 3 moles of 02 this means there is 3*2 = 6 oxygen atoms required

How many moles of Al2O3 are formed?

There are 2 moles of Al2O3 formed

What is the mole ratio of Al to O?

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

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mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b