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When \(1.00 \mathrm{~L}\) of \(2.00 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution at \(30.0^{\circ} \mathrm{C}\) is added to \(2.00 \mathrm{~L}\) of \(0.750 \mathrm{M} \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution at \(30.0^{\circ} \mathrm{C}\) in a calorimeter, a white solid \(\left(\mathrm{BaSO}_{4}\right)\) forms. The temperature of the mixture increases to \(42.0^{\circ} \mathrm{C}\). Assuming that the specific heat capacity of the solution is \(6.37 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that the density of the final solution is \(2.00 \mathrm{~g} / \mathrm{mL}\), calculate the enthalpy change per mole of \(\mathrm{BaSO}_{4}\) formed.

Short Answer

Expert verified
The enthalpy change per mole of BaSO鈧 formed when 1.00 L of 2.00 M Na鈧係O鈧 solution is mixed with 2.00 L of 0.750 M Ba(NO鈧)鈧 solution, and the temperature increases from 30.0 掳C to 42.0 掳C is found to be \(304,896 \ \text{J/mol}\).

Step by step solution

01

1. Write the balanced chemical equation

Before starting any calculations, we should write the balanced chemical equation for the reaction. Na鈧係O鈧 (aq) + Ba(NO鈧)鈧 (aq) 鈫 BaSO鈧 (s) + 2 NaNO鈧 (aq)
02

2. Identify the limiting reactant

To find the limiting reactant, we will compare the mole ratios of the two reactants. The volume and concentration of the reactants are given as: Na鈧係O鈧: 1.00 L 脳 2.00 M = 2.00 moles Ba(NO鈧)鈧: 2.00 L 脳 0.750 M = 1.50 moles The mole ratios from the balanced equation are: Na鈧係O鈧 : Ba(NO鈧)鈧 = 1 : 1 Comparing the mole ratios: 2.00 moles Na鈧係O鈧 / 1 = 2.00 1.50 moles Ba(NO鈧)鈧 / 1 = 1.50 Since 1.50 (Ba(NO鈧)鈧) < 2.00 (Na鈧係O鈧), Ba(NO鈧)鈧 is the limiting reactant.
03

3. Calculate moles of BaSO鈧 formed

As the reaction has a 1:1 mole ratio between Ba(NO鈧)鈧 and BaSO鈧, the moles of BaSO鈧 formed will be equal to the moles of limiting reactant used. Hence: 1.50 moles Ba(NO鈧)鈧 鈫 1.50 moles BaSO鈧
04

4. Calculate the heat released

We need to calculate the heat released (q) during the reaction using the given information. The specific heat capacity (c) is given as 6.37 J/掳C路g, and the density of the final solution (蟻) is 2.00 g/mL. The temperature change (鈭員) experienced by the reaction is: 鈭員 = 42.0掳C - 30.0掳C = 12.0掳C To find the mass (m) of the final solution, we use the given density and total volume (3.00 L): m = 蟻 脳 volume = 2.00 g/mL 脳 3000 mL = 6000 g Now, we can calculate the heat released using the formula: q = mc鈭員 q = (6000 g)(6.37 J/掳C路g)(12.0掳C) q = 457344 J
05

5. Calculate the enthalpy change per mole of BaSO鈧 formed

Now we can calculate the enthalpy change (螖H) per mole of BaSO鈧 by dividing the heat released (q) by the moles of BaSO鈧 formed: 螖H = q / moles of BaSO鈧 螖H = 457344 J / 1.50 moles 螖H = 304896 J/mol So, the enthalpy change per mole of BaSO鈧 formed is 304,896 J/mol.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reactant Determination
Understanding the concept of the limiting reactant is essential for predicting the amount of product that can be formed in a chemical reaction. It refers to the substance that is completely consumed first during the chemical reaction, thus limiting the amount of product that can be formed. This is because no further reaction can occur once the limiting reactant is used up.

To determine the limiting reactant, one must first know the balanced chemical equation, which gives the mole ratios of the reactants and the products. With the mole ratio, you can then compare the number of moles of each reactant available. The reactant that provides the fewest moles of product, according to the stoichiometry of the balanced equation, is the limiting reactant.

In our exercise, the balanced chemical equation is Na鈧係O鈧(aq) + Ba(NO鈧)鈧(aq) 鈫 BaSO鈧(s) + 2 NaNO鈧(aq), and by comparing the available moles of Na鈧係O鈧 and Ba(NO鈧)鈧, we can see that Ba(NO鈧)鈧 is the limiting reactant. This is because even though there are more moles of Na鈧係O鈧, the reaction requires equal amounts of Na鈧係O鈧 and Ba(NO鈧)鈧 for stoichiometric conversion to products.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry provides quantitative information about the proportions in which reactants combine to form products during a chemical reaction. It is derived from the balanced chemical equation, which indicates the ratios of moles of substances involved.

Accurate stoichiometric calculations are fundamental to determine theoretical yields, which represent the maximum amount of product that can be obtained if all of the limiting reactant is converted into product. The stoichiometry of a reaction is based on the law of conservation of mass, stating that matter is neither created nor destroyed during a chemical reaction. Therefore, the amounts of various components need to be balanced.

In the given exercise, we conceptualize this through the equation Na鈧係O鈧(aq) + Ba(NO鈧)鈧(aq) 鈫 BaSO鈧(s) + 2 NaNO鈧(aq), where the stoichiometry (1:1 ratio) informs us that each mole of Ba(NO鈧)鈧 will produce one mole of BaSO鈧. The stoichiometry plays a crucial role in leading us to calculate the correct amount of product formed, which is vital for later determining the enthalpy change per mole.
Heat Capacity and Density in Thermochemistry
In thermochemistry, the concepts of heat capacity and density are important for calculating the amount of heat absorbed or released during a reaction. Heat capacity, defined as the amount of heat energy required to raise the temperature of a substance by one degree Celsius, is crucial when determining the thermal energy change of the system.

Density, the mass per unit volume, is also significant as it helps in determining the mass of solutions or mixtures when volumes are known, as is often the case in chemical reactions. In thermochemical calculations, it is used along with heat capacity to quantify the heat (q) released or absorbed by the reaction, using the formula: q = mc鈭員, where m is the mass, c is the specific heat capacity, and 鈭員 is the change in temperature.

In our example, the specific heat capacity of the solution is 6.37 J/掳C路g, and the density of the final solution is 2.00 g/mL. These values allow us to calculate the total heat change as the product of mass, specific heat capacity, and the temperature difference. This calculation demonstrates how heat capacity and density are applied in practice to determine the enthalpy change for the thermochemical process in question.

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Most popular questions from this chapter

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(1.00 \mathrm{~atm}\) and \(100 .^{\circ} \mathrm{C}\) occupies a volume of \(30.6 \mathrm{~L}\). When one mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to one mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{~g} / \mathrm{cm}^{3}\), calculate \(\Delta E\) for the condensation of one mole of water at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}\).

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm} .\) In the process, there is a heat gain by the system of \(350 . \mathrm{J}\). c. A piston expands against \(1.00 \mathrm{~atm}\) of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

A biology experiment requires the preparation of a water bath at \(37.0^{\circ} \mathrm{C}\) (body temperature). The temperature of the cold tap water is \(22.0^{\circ} \mathrm{C}\), and the temperature of the hot tap water is \(55.0^{\circ} \mathrm{C}\). If a student starts with \(90.0 \mathrm{~g}\) cold water, what mass of hot water must be added to reach \(37.0^{\circ} \mathrm{C} ?\)

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