A sample of 7.75 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 reaction between magnesium hydroxide (Mg(OH)₂) and nitric acid (HNO₃) is an acid-base reaction, forming magnesium nitrate (Mg(NO₃)₂) and water (H₂O) as products. The balanced chemical equation for the reaction is: \[Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O\]

We are given: - Mass of Mg(OH)₂ = 7.75 g - Volume of HNO₃ solution = 25.0 mL - Concentration of HNO₃ solution = 0.200 M Now let's calculate the initial moles of Mg(OH)₂ and HNO₃. (a) Moles of Mg(OH)₂: To calculate moles, we use the formula: \[ moles = \frac{mass}{molar\ mass}\] The molar mass of Mg(OH)₂ is: (24.3 g/mol for Mg) + (2 × (16.0 g/mol for O) + (1.01 g/mol for H)) = 58.3 g/mol So, the moles of Mg(OH)₂: \[moles\ of\ Mg(OH)_2 = \frac{7.75\ g}{58.3\ g/mol} = 0.1329\ mol\] (b) Moles of HNO₃: Since we know the concentration and volume of the HNO₃ solution, we can calculate the moles using the formula: \[moles = concentration \times volume\ (in\ liters)\] The volume of the HNO₃ solution in liters is: \(25.0\ mL = \frac{25.0}{1000}\ L = 0.025\ L\) Now, let's calculate the moles of HNO₃: \[moles\ of\ HNO_3 = 0.200\ M \times 0.025\ L = 0.005\ mol\]

In the balanced chemical equation: \[Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O\] The stoichiometry shows that 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃. We will divide the initial moles of each reactant by its stoichiometric coefficient to determine the limiting reactant: Mg(OH)₂: \(\frac{0.1329\ mol}{1} = 0.1329\ mol\) HNO₃: \(\frac{0.005\ mol}{2} = 0.0025\ mol\) Since 0.0025 mol of HNO₃ is less than 0.1329 mol of Mg(OH)₂, HNO₃ is the limiting reactant.

Since HNO₃ is the limiting reactant, all 0.005 mol of it will be used up in the reaction. The stoichiometry of the reaction shows that 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃, so the change in moles of Mg(OH)₂ is: \[\Delta\ mol\ of\ Mg(OH)_2 = -\frac{0.005\ mol}{2} = -0.0025\ mol\] For Mg(NO₃)₂ and H₂O, we apply the stoichiometry from the balanced equation: \[Mg(OH)_2 + 2HNO_3 \rightarrow Mg(NO_3)_2 + 2H_2O\] \[0.0025\ mol\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.0050\ mol\] So, the moles of Mg(OH)₂, HNO₃, and Mg(NO₃)₂ after the reaction is complete are: Mg(OH)₂: \(0.1329\ mol - 0.0025\ mol = 0.1304\ mol\) HNO₃: \(0.005\ mol - 0.005\ mol = 0\ mol\) Mg(NO₃)₂: \(0 + 0.0025\ mol = 0.0025\ mol\) Thus, after the reaction is complete, there are 0.1304 moles of Mg(OH)₂, 0 moles of HNO₃, and 0.0025 moles of Mg(NO₃)₂ present.