# Stoichiometry: complete precipitation of mercury(II) bromide

I'm really stuck on this question:

What volume of $$0.0712 \mathrm{~mol~dm}^{–3}$$ ammonium bromide solution will be required to precipitate all of the mercury(II) ions from $$33.6 \mathrm{~cm}^3$$ of $$0.144\mathrm{~mol~dm}^{–3}$$ mercury(II) nitrate solution? The ionic equation for the reaction is: $$\ce{Hg^{2+}(aq) + 2Br–(aq) -> HgBr2(s)}$$

The answer given is $$136\mathrm{~mL}$$.

What I have done so far:

I have figured out the number of moles of mercury(II) nitrate which is $$0.05\ \mathrm{mol}$$, and then used this to work out the number of mercury(II) ions ($$3.01\times10^{22}$$). I am unsure where to go from here, would I be required to multiply this value by two to get bromide ions?

1. According to your stoichiometry, how many moles of Br$^-$(aq) ions do you need to react with with one mole of Hg ions?
2. Using that same ratio, how many moles of Br$^-$(aq) ions do you need to react with all of the Hg ions
3. How many moles of Br$^-$(aq) does one mole of NH$_4$Br provide? How many total moles of NH$_4$Br do you need to provide all the moles of Br$^-$(aq) you need?
4. Now that you know how many total moles of NH$_4$Br you need, how can you translate this amount to the total volume required given your concentration of $\frac{0.0712\text{ mol NH}_4\text{Br}}{1 \text{dm}^3}$