Write a balanced chemical equation based on the following description:
solid chromium reacts with solid iodine to form solid chromium(III) iodide.

2 Cr(s) + 3 I2(s) → 2 CrI3(s)

Explanation:

Reactants:

Solid Chromium

• Chromium is a transition metal with an elemental symbol → Cr

• It can exist as a stable monatomic solid → Cr(s)

Solid Iodine

• Iodine is a non-metal halide of Group 7A → I

• It exists as a diatomic solid → I2(s)

Products:

Chromium(III) iodide

• Chromium, Cr → roman numeral III → +3 charge → Cr3+

• Iodide, I → from group 7A → -1 charge → I-

• When ions combine, their charges or oxidation numbers crisscross to become subscripts

• Cr will have a subscript of 1 from I- → Cr

• I will have a subscript of 3 from Cr3+ → I3

• Subscript 1 is no longer indicated → CrI3(s)

Unbalanced Chemical Reaction: Cr(s) + I2(s) → CrI3(s)

Balancing the chemical reaction:

1. Balance I:

Cr(s) + I2(s) → CrI3(s)

• There are 2 I in the reactant and 3 I in the product side

• Balance I by adding the coefficient 3 in I2 and 2 in CrI2

• There would be a total of 6 I in the reactant and product sides

Cr(s) + 3 I2(s) → 2 CrI3(s)

2. Balance Cr:

Cr(s) + 3 I2(s) → 2 CrI3(s)

• There is 1 Cr in the reactant and 2 Cr in the product side

• Balance Cr by adding the coefficient 2 in Cr(s)

• There would be a total of 2 Cr in the reactant and product sides

2 Cr(s) + 3 I2(s) → 2 CrI3(s)

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