II Review | Constants I Periodic Table Butane, C4 H10, reacts with oxygen, O2 , to form water, H2 O, and carbon dioxide, CO2 , as shown in the following chemical equation: Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is 1. Convert from grams of compound X to moles of compound X using the molar mass of compound X. 2. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation. 3. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y. 2C4H10 (g) + 1302(g)→10H2O(g) +8CO2(g) The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react. Also, notice that the coefficient for butane (2) is one- fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth Part B the number of moles of carbon dioxide that you produce. But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios. Calculate the mass of water produced when 6.65 g of butane reacts with excess oxygen. Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) HẢ Value Units Submit Part C Calculate the mass of butane needed to produce 71.4 g of carbon dioxide. Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) ? Value Units Submit
II Review | Constants I Periodic Table Butane, C4 H10, reacts with oxygen, O2 , to form water, H2 O, and carbon dioxide, CO2 , as shown in the following chemical equation: Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is 1. Convert from grams of compound X to moles of compound X using the molar mass of compound X. 2. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation. 3. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y. 2C4H10 (g) + 1302(g)→10H2O(g) +8CO2(g) The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react. Also, notice that the coefficient for butane (2) is one- fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth Part B the number of moles of carbon dioxide that you produce. But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios. Calculate the mass of water produced when 6.65 g of butane reacts with excess oxygen. Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) HẢ Value Units Submit Part C Calculate the mass of butane needed to produce 71.4 g of carbon dioxide. Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) ? Value Units Submit
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...
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![**Butane Reaction with Oxygen**
**Introduction:**
Butane, \( \text{C}_4\text{H}_{10} \), reacts with oxygen, \( \text{O}_2 \), to form water, \( \text{H}_2\text{O} \), and carbon dioxide, \( \text{CO}_2 \), as shown in the following chemical equation:
\[ 2\text{C}_4\text{H}_{10} (g) + 13\text{O}_2 (g) \rightarrow 10\text{H}_2\text{O} (g) + 8\text{CO}_2 (g) \]
**Understanding Coefficients:**
The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react.
Additionally, the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce.
**Important Note:**
Be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios.
**Stoichiometry Strategy:**
Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is:
1. Convert from grams of compound X to moles of compound X using the molar mass of compound X.
2. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation.
3. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y.
**Exercises:**
**Part B:**
*Calculate the mass of water produced when 6.65 g of butane reacts with excess oxygen.*
- Express your answer to three significant figures and include the appropriate units.
- [View Available Hint(s)]
**Value Input Section:**
- [Input Box] | Units
**Part C:**
*Calculate the mass of butane needed to produce 71.4 g of carbon dioxide.*
- Express your answer to three significant figures and include the appropriate units.
- [View Available](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbcea870b-a558-4e25-bd85-f50b869e1110%2F36dcc576-9ef6-430a-a3ca-099a00fead77%2Friemo1b_processed.png&w=3840&q=75)
Transcribed Image Text:**Butane Reaction with Oxygen**
**Introduction:**
Butane, \( \text{C}_4\text{H}_{10} \), reacts with oxygen, \( \text{O}_2 \), to form water, \( \text{H}_2\text{O} \), and carbon dioxide, \( \text{CO}_2 \), as shown in the following chemical equation:
\[ 2\text{C}_4\text{H}_{10} (g) + 13\text{O}_2 (g) \rightarrow 10\text{H}_2\text{O} (g) + 8\text{CO}_2 (g) \]
**Understanding Coefficients:**
The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react.
Additionally, the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce.
**Important Note:**
Be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios.
**Stoichiometry Strategy:**
Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is:
1. Convert from grams of compound X to moles of compound X using the molar mass of compound X.
2. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation.
3. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y.
**Exercises:**
**Part B:**
*Calculate the mass of water produced when 6.65 g of butane reacts with excess oxygen.*
- Express your answer to three significant figures and include the appropriate units.
- [View Available Hint(s)]
**Value Input Section:**
- [Input Box] | Units
**Part C:**
*Calculate the mass of butane needed to produce 71.4 g of carbon dioxide.*
- Express your answer to three significant figures and include the appropriate units.
- [View Available
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