A sample of \(5.53 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is added to \(25.0 \mathrm{~mL}\) of 0.200 \(\mathrm{M} \mathrm{HNO}_{3}\) (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Mg}(\mathrm{OH})_{2}, \mathrm{HNO}_{3},\) and \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) are present after the reaction is complete?

The balanced chemical equation for the neutralization reaction between magnesium hydroxide and nitric acid is given by: \( Mg(OH)_2 + 2 HNO_3 \rightarrow Mg(NO_3)_2 + 2 H_2O \) This equation indicates that for every mole of magnesium hydroxide reacted, two moles of nitric acid are also reacted, forming one mole of magnesium nitrate and two moles of water.

First, we need to calculate the moles of magnesium hydroxide and nitric acid present in the reaction. Moles of Mg(OH)₂ = mass / molar mass The molar mass of Mg(OH)₂ = 24.31 (Mg) + 2 * (16.00 (O) + 1.01 (H)) = 58.33 g/mol Moles of Mg(OH)₂ = 5.53 g / 58.33 g/mol ≈ 0.0948 mol Moles of HNO₃ = concentration * volume To convert the volume from milliliters to liters: 25.0 mL * (1L/1000mL) = 0.025 L Moles of HNO₃ = (0.200 mol/L) * 0.025 L = 0.00500 mol

To find out which reactant is the limiting reactant, we compare the moles ratios from the balanced chemical equation to the initial moles we have: Mg(OH)₂ : HNO₃ required for reaction = 1 : 2 Now we must compare the ratio of initial moles of Mg(OH)₂ and HNO₃ to this mol ratio from the balanced equation: 0.0948 mol Mg(OH)₂ : 0.00500 mol HNO₃ We can see that there are not enough moles of HNO₃ relative to Mg(OH)₂. Thus, HNO₃ is the limiting reactant.

As HNO₃ is the limiting reactant, we use its moles to determine the amounts of the other compounds: (a) Mg(OH)₂ From the stoichiometry, 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃. Therefore, Moles of Mg(OH)₂ reacted = Moles of HNO₃ / 2 = 0.00500 mol / 2 = 0.00250 mol Moles of Mg(OH)₂ remaining = initial moles - reacted moles = 0.0948 mol - 0.00250 mol ≈ 0.0923 mol (b) HNO₃ Since HNO₃ is the limiting reactant, all of its moles are consumed by the reaction. Moles of HNO₃ remaining = 0 mol (c) Mg(NO₃)₂ From the stoichiometry, 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃ to form 1 mole of Mg(NO₃)₂. Thus, the moles of Mg(NO₃)₂ formed are equal to the moles of Mg(OH)₂ reacted. Moles of Mg(NO₃)₂ formed = 0.00250 mol In summary, after the reaction is complete, there are approximately 0.0923 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.00250 moles of Mg(NO₃)₂.