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Chapter 4: Problem 109

A sample of \(8.69 \mathrm{~g}\) of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is added to \(155.0 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .\) (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of \(\mathrm{Zn}(\mathrm{OH})_{2}, \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(\mathrm{ZnSO}_{4}\) are present after the reaction is complete?

Short Answer

Expert verified
The balanced chemical equation for the reaction is: \( Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O \). Zn(OH)₂ is the limiting reactant. After the reaction is complete, there are 0 moles of \( Zn(OH)_2 \), 0.0289 moles of \( H_2SO_4 \), and 0.0874 moles of \( ZnSO_4 \).

Step by step solution

01

Write the balanced chemical equation

First, we have to determine the products of the reaction between Zn(OH)₂ and H₂SO₄. Zn(OH)₂ is a metal hydroxide while H₂SO₄ is an acid, so based on the acid-base reaction, the products are salt and water. The balanced chemical equation: \( Zn(OH)_2 + H_2SO_4 \rightarrow ZnSO_4 + 2H_2O \)
02

Find the moles of given reactants

We need to find the number of moles of Zn(OH)₂ and H₂SO₄ in the given amounts, using the formula mass(molar_mass) and molarity. For Zn(OH)₂: The molar mass of Zn = 65.38 g/mol The molar mass of O = 16.00 g/mol The molar mass of H = 1.01 g/mol The formula mass of Zn(OH)₂ = (65.38 + 2 * (16.00 + 1.01)) = 65.38 + 2 * 17.01 = 99.40 g/mol From the given amount of Zn(OH)₂ (8.69 g), the number of moles = 8.69 g ÷ 99.40 g/mol = 0.0874 mol For H₂SO₄: The given molarity of H₂SO₄ = 0.750 M, which means 0.750 moles are present in 1 L of solution. So, in 155 mL (0.155 L), the number of moles =0.750 mol * 0.155 L = 0.1163 mol
03

Identify the limiting reactant

We can use the stoichiometry of the balanced chemical equation to calculate the required moles of H₂SO₄ for the given moles of Zn(OH)₂: \( Required~moles~of~H_2SO_4 = 1 * 0.0874~mol~Zn(OH)_2 = 0.0874~ mol~H_2SO_4 \) As we have 0.1163 mol of H₂SO₄, it is more than required, so Zn(OH)₂ is the limiting reactant.
04

Calculate the moles of reactants and products after the reaction

As Zn(OH)₂ is the limiting reactant, all of it will be consumed, so there will be 0 moles of Zn(OH)₂ left after the reaction. Moles of H₂SO₄ consumed = moles of Zn(OH)₂ = 0.0874 mol Remaining moles of H₂SO₄ = 0.1163 - 0.0874 = 0.0289 mol Based on the stoichiometry of the balanced equation, 1 mol of Zn(OH)₂ reacts with 1 mol of H₂SO₄ to produce 1 mol of ZnSO₄. So, the moles of ZnSO₄ produced = moles of Zn(OH)₂ consumed = 0.0874 mol In summary, after the reaction is complete, we have: - moles of Zn(OH)₂ = 0 mol - moles of H₂SO₄ = 0.0289 mol - moles of ZnSO₄ = 0.0874 mol

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Most popular questions from this chapter

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{Cl}^{-}\) ion: (a) \(0.10 \mathrm{MAlCl}_{3}\) solution or a $0.25 \mathrm{MLiCl}\( solution, (b) \)150 \mathrm{~mL}\(. of a \)0.05 \mathrm{MMnCl}_{3}\( solution or \)200 \mathrm{~mL}$. of \(0.10 \mathrm{M} \mathrm{KCl}\) solution, (c) a \(2.8 M H C l\) solution or a solution made by dissolving $23.5 \mathrm{~g}\( of \)\mathrm{KCl}\( in water to make \)100 \mathrm{~mL}$ of solution.

Using the activity series (Table 4.5 ), write balanced chemical equations for the following reactions. If no reaction occurs, write NR. (a) Iron metal is added to a solution of copper(II) nitrate, (b) zinc metal is added to a solution of magnesium sulfate, (c) hydrobromic acid is added to tin metal, (d) hydrogen gas is bubbled through an aqueous solution of nickel(II) chloride, (e) aluminum metal is added to a solution of cobalt(II) sulfate.

(a) You have a stock solution of \(14.8 \mathrm{M} \mathrm{NH}_{3}\). How many milliliters of this solution should you dilute to make \(1000.0 \mathrm{~mL}\) of \(0.250 \mathrm{MNH}_{3} ?\) (b) If you take a \(10.0-\mathrm{mL}\) portion of the stock solution and dilute it to a total volume of \(0.500 \mathrm{~L},\) what will be the concentration of the final solution?

Which of the following ions will always be a spectator ion in a precipitation reaction? (a) $\mathrm{Cl}^{-},(\mathbf{b}) \mathrm{NO}_{3}^{-}$, (d) \(\mathrm{S}^{2-}\), (c) \(\mathrm{NH}_{4}^{+}\) (e) \(\mathrm{SO}_{4}^{2-} .[\) Section 4.2\(]\)

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of $0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}$ solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH}\) ? (c) If $55.8 \mathrm{~mL}\( of a \)\mathrm{BaCl}_{2}$ solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg}\) sample of $\mathrm{Na}_{2} \mathrm{SO}_{4}\(, what is the molarity of the \)\mathrm{BaCl}_{2}$ solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{MHCl}\) solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\), how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?
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