Chapter 6: Problem 27
What is incomplete combustion of fossil fuels? Why can this be a problem?
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 6: Problem 27
What is incomplete combustion of fossil fuels? Why can this be a problem?
All the tools & learning materials you need for study success - in one app.
Get started for free
Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{L}\) at 1.0 atm and \(25^{\circ} \mathrm{C} .\) How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?
A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.
The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)(\text { unbalanced }) $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{kJ}\) . If \(2.50 \times 10^{2} \mathrm{mL} \mathrm{N}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and 3.50 atm and \(4.50 \times\) \(10^{2} \mathrm{mL} \mathrm{O}_{2}(g)\) at \(100 .^{\circ} \mathrm{C}\) and 3.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)
A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing 150.0 \(\mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\) . After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C}\) . If the specific heat capacity of nickel is 0.444 \(\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.
Explain why \(\Delta H\) is obtained directly from coffee-cup calorimeters, whereas \(\Delta E\) is obtained directly from bomb calorimeters.
What do you think about this solution?
We value your feedback to improve our textbook solutions.