Write down the following equation in words Fe+aO² Fe²O³​

4Fe+3O2→2Fe2O3

Step 1: Tally the number of atoms based on subscript.

F e + O 2 → F e 2 O 3 (unbalanced) left side: Fe = 1; O = 2 right side: Fe = 2; O = 3 Step 2: Start balancing the simple molecules first, in this case the O 2 F e + 3 O 2 → F e 2 O 3 left side: Fe = 1; O = 2 x 3 = 6 right side: Fe = 2; O = 3 Since there are now 6 O atoms on the left, there must also be 6 O atoms on the right. Notice that the right hand is one big molecule so whatever coefficient you would use to balance the O atoms, you would also need to apply to the bonded F e atom. F e + 3 O 2 → 2 F e 2 O 3 left side: Fe = 1; O = 2 x 3 = 6 right side: Fe = 2 x 2 = 4; O = 3 x 2 = 6 Step 3: Now balance the rest of the equation. 4 F e + 3 O 2 → 2 F e 2 O 3 left side: Fe = 1 x 4 = 4; O = 2 x 3 = 6 right side: Fe = 2 x 2 = 4; O = 3 x 2 = 6 The equation is now balanced.

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

dynamic equilibrium is the answer.