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Chapter 4: Problem 109

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

Expert verified
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

01

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: \(Mg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O\)
02

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: \(moles = Molarity × Volume(in~Liters)\) \(moles~of~HNO₃ = 0.200~M \times 0.0250~L = 0.00500~mol\)
03

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.
04

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.

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Most popular questions from this chapter

(a) Calculate the molarity of a solution that contains \(0.0250 \mathrm{~mol} \mathrm{NH}_{4} \mathrm{Cl}\) in exactly \(500 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{HNO}_{3}\) are present in \(50.0 \mathrm{~mL}\) of a \(2.50 \mathrm{M}\) solution of nitric acid? (c) How many milliliters of \(1.50 \mathrm{M} \mathrm{KOH}\) solution are needed to provide \(0.275 \mathrm{~mol}\) of KOH?

(a) Calculate the molarity of a solution made by dissolving \(0.750\) grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in enough water to form exactly \(850 \mathrm{~mL}\) of solution. (b) How many moles of \(\mathrm{KMnO}_{4}\) are present in \(250 \mathrm{~mL}\) of a \(0.0475 \mathrm{M}\) solution? (c) How many milliliters of \(11.6 \mathrm{M} \mathrm{HCl}\) solution are needed to obtain \(0.250 \mathrm{~mol}\) of \(\mathrm{HCl}\) ?

(a) How many milliliters of a stock solution of \(10.0 \mathrm{M}\) \(\mathrm{HNO}_{3}\) would you have to use to prepare \(0.450 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{HNO}_{3} ?(\mathrm{~b})\) If you dilute \(25.0 \mathrm{~mL}\) of the stock so- lution to a final volume of \(0.500 \mathrm{~L}\), what will be the concentration of the diluted solution?

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodium bicarbonate reacts with sulfuric acid as follows: \(2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) $$ Sodium bicarbonate is added untilthe fizzing due to the formation of \(\mathrm{CO}_{2}(\mathrm{~g})\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\) that must be added to the spill to neutralize the acid?

Explain the following observations: (a) \(\mathrm{NH}_{3}\) contains no \(\mathrm{OH}^{-}\) ions, and yet its aqueous solutions are basic; (b) HF is called a weak acid, and yet it is very reactive; (c) although sulfuric acid is a strong electrolyte, an aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) contains more \(\mathrm{HSO}_{4}^{-}\) ions than \(\mathrm{SO}_{4}^{2-}\) ions.
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