Question

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?

Short answer

The balanced chemical equation for the reaction between magnesium hydroxide (Mg(OH)₂) and nitric acid (HNO₃) is as follows: \(Mg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O\) In this reaction, the limiting reactant is HNO₃. After the reaction is complete, there are 0.0923 mol of Mg(OH)₂, 0 mol of HNO₃, and 0.00250 mol of Mg(NO₃)₂ present.

Step-by-step solution

Write the chemical equation for the reaction

The reaction between magnesium hydroxide (Mg(OH)₂) and nitric acid (HNO₃) produces magnesium nitrate (Mg(NO₃)₂) and water (H₂O). The balanced chemical equation for this reaction is: \(Mg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O\)

Calculate moles of reactants

We are given 5.53 g of Mg(OH)₂ and 25.0 mL of 0.200 M HNO₃. First, we will convert the given amounts to moles. (a) Moles of Mg(OH)₂: \(moles = \frac{mass}{molar~mass}\) The molar mass of Mg(OH)₂ is 24.31 g/mol (Mg) + 34.02 g/mol (2 × OH) = 58.33 g/mol. \(moles~of~Mg(OH)₂ = \frac{5.53~g}{58.33~g/mol} = 0.0948~mol\) (b) Moles of HNO₃: \(moles = Molarity × Volume(in~Liters)\) \(moles~of~HNO₃ = 0.200~M \times 0.0250~L = 0.00500~mol\)

Determine the limiting reactant

According to the balanced equation, 1 mole of Mg(OH)₂ reacts with 2 moles of HNO₃. We will find out how many moles of HNO₃ are needed to react with the given moles of Mg(OH)₂ and compare it with the available moles of HNO₃. Moles of HNO₃ needed to react with 0.0948 mol of Mg(OH)₂: \(moles~of~HNO₃ = 2 \times moles~of~Mg(OH)₂\) \(moles~of~HNO₃ = 2 × 0.0948~mol = 0.1896~mol\) Since we have only 0.00500 mol of HNO₃ available, HNO₃ is the limiting reactant.

Calculate moles of Mg(OH)₂, HNO₃, and Mg(NO₃)₂ after the reaction

(a) Since HNO₃ is the limiting reactant, it will be completely used up: Moles of HNO₃ left = 0 (b) Calculate the amount of Mg(OH)₂ that reacted: \(moles~of~Mg(OH)₂~reacted = \frac{1}{2} \times moles~of~HNO₃\) \(moles~of~Mg(OH)₂~reacted = \frac{1}{2} × 0.00500~mol = 0.00250~mol\) Moles of Mg(OH)₂ left = 0.0948 - 0.00250 = 0.0923 mol (c) Calculate the moles of Mg(NO₃)₂ produced: \(moles~of~Mg(NO₃)₂ = \frac{1}{2} \times moles~of~HNO₃\) \(moles~of~Mg(NO₃)₂ = \frac{1}{2} × 0.00500~mol = 0.00250~mol\) After the reaction is complete, there are 0.0923 mol of Mg(OH)₂, 0 mol of HNO₃, and 0.00250 mol of Mg(NO₃)₂ present.