Finding half-life of a nuclide based on gas release Full Question: The hydride of an unstable nuclide of a Group IIA metal, $\ce{MH2 (s)}$, decays by alpha emission. A $\pu{0.025 mol}$ sample of the hydride is placed in an evacuated $\pu{2.0 L}$ container at $\pu{298 K}$. After 82 minutes, the pressure in the container is $\pu{0.55 atm}$. Find the half-life of the nuclide.
Here's what I have so far:

Find nuclear equation MH2(s) ---&gt; He(g) + Noble gas + H2 (g) (not sure if this is right)


Find total moles of gas in container at 80 mins
$$n = \frac{PV}{RT} = \frac{(\pu{0.55 atm})(\pu{2 L})}{(\pu{0.0821 L atm mol-1 K-1})(\pu{298 K})} = \pu{0.045 mol}$$


Use integrated first rate law to calculate k and half-life.
Use $\ln N_t = -kt + \ln N_i$, where $N_i$ is the initial amount of moles of $\ce{MH2}$ ($\pu{0.025 mol}$) and $N_t$ is the moles of hydride remaining after 80 minutes.


Then find half life using k.
$$t_{1/2} = \ln 2 / k$$

Here's what I'm having trouble on: How to find $N_t$, i.e. the moles of metal hydride remaining. This part has been stumping me for quite some time, so help is appreciated! Thank you!
For this we can first write the equation for the reaction again:
$$MH_2 =&gt; He_{(g)} + Noble gas + H_{2(g)}$$
Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then we can use the first order integrated rate law now:
$$ln(N_f) = -kt + ln(N_i)$$
$$ln(0.025-0.015) = -k(80 mins) + ln(0.025)$$
$$k = 0.0115 min^{-1}$$
Now that we have k, we can solve for the half-life.
$$t_{1/2}=\frac{ln(2)}{k}$$
$$t_{1/2}=\frac{ln(2)}{0.0115min^{-1}} \approx 61 mins$$