Question

1. write a balanced chemical equation for the reaction of solid strontium with iodine gas.

Explanation:

Step1: Identify reactants and products

Reactants: Solid strontium ($\text{Sr}(s)$) and iodine gas ($\text{I}_2(g)$). Product: Strontium iodide. Strontium is in group 2, so it forms a +2 ion ($\text{Sr}^{2+}$). Iodine is in group 17, so it forms a -1 ion ($\text{I}^-$). To form a neutral compound, the formula for strontium iodide is $\text{SrI}_2$.

Step2: Write the unbalanced equation

$\text{Sr}(s) + \text{I}_2(g)
ightarrow \text{SrI}_2(s)$

Step3: Check for balance

Count the number of each atom on both sides.

• Strontium (Sr): 1 on left, 1 on right.
• Iodine (I): 2 on left, 2 on right.

The equation is already balanced.

Answer:

$\text{Sr}(s) + \text{I}_2(g)
ightarrow \text{SrI}_2(s)$