The decay of Iridium-192 which is a first-order reaction with a half-life of 74 days?

a.
The balanced equation is shown at the bottom.

The rate equation:  Rate = k A,
where A is the amount of iridium-192.

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b.
Rate constant, k = ln(2)/t½ = ln(2)/(74 days) = 0.009367 day⁻¹ ≈ 0.00937 day⁻¹

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c.
No. of half-lives = (5 × 365.25 days) / (74 days) = 24.68
Mass after 5 years = (50.0 g) × (1/2)²⁴·⁶⁸ = 1.86 × 10⁻⁶ g

OR:
Time taken, t = 5 × 365.25 days = 1826.25 days
Mass after 5 years, A = Aₒ e⁻ᵏᵗ = (50.0 g) × e⁻⁰·⁰⁰⁹³⁶⁷ˣ¹⁸²⁶·²⁵ = 1.86 × 10⁻⁶ g