Answered: Hydrocarbon mixtures are used as fuels.…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

Hydrocarbon mixtures are used as fuels. (a) How manygrams of CO₂(g) are produced by the combustion of 200. g of amixture that is 25.0% CH₄ and 75.0% C₃H₈ by mass? (b) A 252-ggaseous mixture of CH₄ and C₃H₈ burns in excess O₂, and 748 g of CO₂ gas is collected. What is the mass % of CH₄ in the mixture?

Expert Solution

Step 1

In a balanced chemical reaction, stoichiometry gives the relationship between the amount of reactants and products. It can be used to calculate the amount of reactant or product by using the amount of product or reactant involved in the balanced reaction respectively.

Step 2

a. Mass percentage of $C H_{4}$ is $25.0 %$ and the mass percentage of $C_{3} H_{8}$ is $75.0 %$ . Therefore, the mass of $C H_{4}$ in $200 g$ compound is as follows:

$\begin{array}{rcl} M a s s o f C H_{4} & = & (\frac{25 %}{100 %}) (200 g) \\ = & 50 g \end{array}$

Hence, the mass of $C_{3} H_{8}$ in $200 g$ compound is as follows:

$\begin{array}{rcl} M a s s o f C H_{4} & = & (\frac{75 %}{100 %}) (200) \\ = & 150 g \end{array}$

The formula to calculate the moles is as follows:

$M o l e s = \frac{M a s s}{M o l a r M a s s}$

Substitute $50 g$ for mass and $16.04 g / m o l$ for molar mass to calculate the moles of $C H_{4}$ .

$\begin{array}{rcl} {(M o l e s)}_{C H_{4}} & = & \frac{50 g}{16.04 g / m o l} \\ = & 3.11720698 m o l \end{array}$

Step 3

The balanced combustion reaction of $C H_{4}$ is as follows:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$

According to the balanced reaction, one mole of $C H_{4}$ produces one mole of $C O_{2}$ and therefore $\begin{array}{rcl} 3.11720698 m o l \end{array}$ of $C H_{4}$ produces $\begin{array}{rcl} 3.11720698 m o l \end{array}$ of $C O_{2}$ .

Substitute $150 g$ for mass and $44.1 g / m o l$ for molar mass to calculate the moles of $C_{3} H_{8}$ .

$\begin{array}{rcl} {(M o l e s)}_{C_{3} H_{8}} & = & \frac{150 g}{44.1 g / m o l} \\ = & 3.40136054 m o l \end{array}$

Step 4

The balanced combustion reaction of $C_{3} H_{8}$ is as follows:

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O$

According to the balanced reaction, one mole of $C_{3} H_{8}$ produces three moles of $C O_{2}$ and therefore $3.40136054 m o l$ of $C_{3} H_{8}$ produces $10.2040816 m o l$ of $C O_{2}$ .

The total moles of $C O_{2}$ is the sum of moles of $C O_{2}$ produced from $C H_{4}$ and moles of $C O_{2}$ produced from $C_{3} H_{8}$ .

$\begin{array}{rcl} T o t a l M o l e s o f C O_{2} & = & 3.11720698 m o l + 10.2040816 m o l \\ = & 13.3212886 m o l \end{array}$

Step 5

The formula to calculate the mass of $C O_{2}$ is as follows:

$M a s s o f C O_{2} = (M o l e s) (M o l a r M a s s)$

Substitute $13.3212886 m o l$ for moles and $44.01 g / m o l$ for molar mass to calculate the mass of $C O_{2}$ .

$\begin{array}{rcl} M a s s o f C O_{2} & = & (13.3212886 m o l) (44.01 g / m o l) \\ = & 586 g \end{array}$

Hence, the mass of $C O_{2}$ is $\begin{array}{rcl} 586 g \end{array}$ .

Step by step

Solved in 9 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions