Answer
The limiting reactant is $HNO_3$.
Work Step by Step
1. Find the nº of moles ($Mg(OH)_2$):
Molar Mass ($Mg(OH)_2$):
Mg: 24.3
O: 16 * 2 = 32
H: 1 * 2 = 2
24.3 + 32 + 2 = 58.3g/mol
$n(moles) = \frac{mass(g)}{mm}$
$n(moles) = \frac{7.75}{58.3} \approx 0.133$
2. Find the nº of moles ($HNO_3$):
$n(moles) = Concentration * Volume(L)$
$n(moles) = 0.005$
3. Find the nº of moles $(HNO_3)$ needed to react with 0.133 moles of $Mg(OH)_2$
Since the proportion is 2 $HNO_3$ to 1 $Mg(OH)_2$
$\frac{1}{2} = \frac{0.133}{x}$
$x = 0.266$
Because there are just 0.005 moles, $HNO_3$ is the limiting reactant.
