Question

1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?

Answer 1

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.