QuestionMay 2, 2025

1. Given the following reaction: underline ( )C_(3)H_(8)+underline ( )O_(2)arrow underline ( )CO_(2)+underline ( )H_(2)O a. Balance the equation first b. Determine the limiting reagent if you start with 14.8 g of C_(3)H_(8) and 3.44 g of O_(2) c. Determine the number of moles of carbon dioxide produced d. Determine the number of grams of H_(2)O produced e. Determine the number of grams of excess reagent left over after the reaction.

1. Given the following reaction: underline ( )C_(3)H_(8)+underline ( )O_(2)arrow underline ( )CO_(2)+underline ( )H_(2)O a. Balance the equation first b. Determine the limiting reagent if you start with 14.8 g of C_(3)H_(8) and 3.44 g of O_(2) c. Determine the number of moles of carbon dioxide produced d. Determine the number of grams of H_(2)O produced e. Determine the number of grams of excess reagent left over after the reaction.
1. Given the following reaction: underline ( )C_(3)H_(8)+underline ( )O_(2)arrow underline ( )CO_(2)+underline ( )H_(2)O a. Balance the equation first b. Determine the limiting reagent if you start with 14.8 g of C_(3)H_(8) and 3.44 g of O_(2) c. Determine the number of moles of carbon dioxide produced d. Determine the number of grams of H_(2)O produced e. Determine the number of grams of excess reagent left over after the reaction.

Solution
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Answer

a. Balanced Equation: C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O ### b. Limiting Reagent: O_{2} ### c. Moles of CO_{2} produced: 0.0645 mol ### d. Grams of H_{2}O produced: 1.548 g ### e. Grams of excess C_{3}H_{8} left: 13.82 g Explanation 1. Balance the chemical equation The balanced equation is C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O. 2. Calculate moles of C_{3}H_{8} and O_{2} Molar mass of C_{3}H_{8} is 44.1 g/mol. Moles of C_{3}H_{8} = \frac{14.8}{44.1} = 0.335 mol. Molar mass of O_{2} is 32.0 g/mol. Moles of O_{2} = \frac{3.44}{32.0} = 0.1075 mol. 3. Determine the limiting reagent From the balanced equation, 1 mol of C_{3}H_{8} reacts with 5 mol of O_{2}. Therefore, 0.335 mol of C_{3}H_{8} requires 0.335 \times 5 = 1.675 mol of O_{2}. Since only 0.1075 mol of O_{2} is available, O_{2} is the limiting reagent. 4. Calculate moles of CO_{2} produced From the balanced equation, 5 mol of O_{2} produces 3 mol of CO_{2}. Therefore, 0.1075 mol of O_{2} produces \frac{3}{5} \times 0.1075 = 0.0645 mol of CO_{2}. 5. Calculate grams of H_{2}O produced From the balanced equation, 5 mol of O_{2} produces 4 mol of H_{2}O. Therefore, 0.1075 mol of O_{2} produces \frac{4}{5} \times 0.1075 = 0.086 mol of H_{2}O. Molar mass of H_{2}O is 18.0 g/mol. Grams of H_{2}O = 0.086 \times 18.0 = 1.548 g. 6. Calculate grams of excess C_{3}H_{8} left over Moles of C_{3}H_{8} reacted = \frac{0.1075}{5} = 0.0215 mol. Moles of C_{3}H_{8} remaining = 0.335 - 0.0215 = 0.3135 mol. Grams of C_{3}H_{8} remaining = 0.3135 \times 44.1 = 13.82 g.

Explanation

1. Balance the chemical equation<br /> The balanced equation is $C_{3}H_{8} + 5O_{2} \rightarrow 3CO_{2} + 4H_{2}O$.<br /><br />2. Calculate moles of $C_{3}H_{8}$ and $O_{2}$<br /> Molar mass of $C_{3}H_{8}$ is 44.1 g/mol. Moles of $C_{3}H_{8}$ = $\frac{14.8}{44.1} = 0.335$ mol.<br /> Molar mass of $O_{2}$ is 32.0 g/mol. Moles of $O_{2}$ = $\frac{3.44}{32.0} = 0.1075$ mol.<br /><br />3. Determine the limiting reagent<br /> From the balanced equation, 1 mol of $C_{3}H_{8}$ reacts with 5 mol of $O_{2}$. Therefore, 0.335 mol of $C_{3}H_{8}$ requires $0.335 \times 5 = 1.675$ mol of $O_{2}$.<br /> Since only 0.1075 mol of $O_{2}$ is available, $O_{2}$ is the limiting reagent.<br /><br />4. Calculate moles of $CO_{2}$ produced<br /> From the balanced equation, 5 mol of $O_{2}$ produces 3 mol of $CO_{2}$. Therefore, 0.1075 mol of $O_{2}$ produces $\frac{3}{5} \times 0.1075 = 0.0645$ mol of $CO_{2}$.<br /><br />5. Calculate grams of $H_{2}O$ produced<br /> From the balanced equation, 5 mol of $O_{2}$ produces 4 mol of $H_{2}O$. Therefore, 0.1075 mol of $O_{2}$ produces $\frac{4}{5} \times 0.1075 = 0.086$ mol of $H_{2}O$.<br /> Molar mass of $H_{2}O$ is 18.0 g/mol. Grams of $H_{2}O$ = $0.086 \times 18.0 = 1.548$ g.<br /><br />6. Calculate grams of excess $C_{3}H_{8}$ left over<br /> Moles of $C_{3}H_{8}$ reacted = $\frac{0.1075}{5} = 0.0215$ mol.<br /> Moles of $C_{3}H_{8}$ remaining = $0.335 - 0.0215 = 0.3135$ mol.<br /> Grams of $C_{3}H_{8}$ remaining = $0.3135 \times 44.1 = 13.82$ g.
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