Butane lighters work by the release and combustion of pressurised butane: 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l Determine the limiting reagent in the reaction, 20 molecules of C 4 H 10 and 100 molecules of O 2 .

Consider the given reaction, 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l . We know that, one mole of C 4 H 10 contains 6 . 02 × 10 23 molecules. Therefore, the number of moles of butane in 20 molecules = 20 6 . 02 × 10 23 = 3 . 32 × 10 - 23 mol , and The number of moles of oxygen in 100 molecules = 100 6 . 02 × 10 23 = 16 . 61 × 10 - 23 mol According to the balanced equation, 2 mol of butane requires 13 mol of oxygen. Therefore, the number of moles of oxygen required to react with 3 . 32 × 10 - 23 mol = 3 . 32 × 10 - 23 × 13 2 = 21 . 58 × 10 - 23 mol . But, only 16 . 61 × 10 - 23 mol of oxygen is available in the reaction, hence, oxygen is the limiting reagent in the reaction.

Butane lighters work by the release and combustion of pressurised butane: 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l Determine the limiting reagent in the reaction, 10 molecules of C 4 H 10 and 91 molecules of O 2 .

Consider the given reaction, 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l . We know that, one mole of C 4 H 10 contains 6 . 02 × 10 23 molecules. Therefore, the number of moles of butane in 10 molecules = 10 6 . 02 × 10 23 = 1 . 66 × 10 - 23 mol , and The number of moles of oxygen in 91 molecules = 91 6 . 02 × 10 23 = 15 . 11 × 10 - 23 mol According to the balanced equation, 2 mol of butane requires 13 mol of oxygen. Therefore, the number of moles of oxygen required to react with 1 . 66 × 10 - 23 mol = 1 . 66 × 10 - 23 × 13 2 = 10 . 79 × 10 - 23 mol . But, 15 . 11 × 10 - 23 mol of oxygen is available in the reaction, hence, butane is the limiting reagent in the reaction.

Butane lighters work by the release and combustion of pressurised butane: 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l Determine the limiting reagent in the reaction, 0 . 20 mol of C 4 H 10 and 2 . 6 mol of O 2 .

Consider the given reaction, 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l . According to the balanced equation, 2 mol of butane requires 13 mol of oxygen. Therefore, the number of moles of oxygen required to react with 0 . 20 mol of butane = 0 . 20 × 13 2 = 1 . 3 mol . But, 2 . 6 mol of oxygen is available in the reaction, hence, butane is the limiting reagent in the reaction.

Butane lighters work by the release and combustion of pressurised butane: 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l Determine the limiting reagent in the reaction, 8 . 72 g of C 4 H 10 and 28 . 8 g of O 2 .

Consider the given reaction, 2 C 4 H 10 g + 13 O 2 g → 8 CO 2 g + 10 H 2 O l . The amount of butane (in mol ) = 8 . 72 58 = 0 . 15 mol The amount of oxygen (in mol ) = 28 . 8 32 = 0 . 9 mol According to the balanced equation, 2 mol of butane requires 13 mol of oxygen. Therefore, the number of moles of oxygen required to react with 0 . 15 mol of butane = 0 . 15 × 13 2 = 0 . 975 mol . But, only 0 . 9 mol of oxygen is available in the reaction, hence, oxygen is the limiting reagent in the reaction.

Chlorine gas is produced by the reaction of hydrochloric acid, and the oxidising agent manganese( IV ) oxide, MnO 2 : MnO 2 s + 4 HCl aq → MnCl 2 aq + Cl 2 g + 2 H 2 O l At 273 . 15 K and 100 kPa , 58 . 34 g of HCl reacts with 0 . 35 mol of MnO 2 to produce 7 . 056 dm 3 of chlorine gas. Deduce the limiting reagent.

Consider the given reaction, MnO 2 s + 4 HCl aq → MnCl 2 aq + Cl 2 g + 2 H 2 O l . The amount of hydrochloric acid (in mol ) = 58 . 34 36 . 5 = 1 . 598 mol According to the balanced equation, 1 mol of MnO 2 reacts with 4 mol of hydrochloric acid. Therefore, the number of moles of hydrochloric acid required to react with 0 . 35 mol of MnO 2 = 0 . 35 × 4 = 1 . 4 mol . But, 1 . 598 mol of hydrochloric acid is available in the reaction (i.e., HCl is in excess) hence, MnO 2 is the limiting reagent in the reaction.

Chlorine gas is produced by the reaction of hydrochloric acid, and the oxidising agent manganese( IV ) oxide, MnO 2 : MnO 2 s + 4 HCl aq → MnCl 2 aq + Cl 2 g + 2 H 2 O l At 273 . 15 K and 100 kPa , 58 . 34 g of HCl reacts with 0 . 35 mol of MnO 2 to produce 7 . 056 dm 3 of chlorine gas. Calculate the theoretical yield of chlorine.

Consider the given reaction, MnO 2 s + 4 HCl aq → MnCl 2 aq + Cl 2 g + 2 H 2 O l . According to the balanced equation, 1 mol of MnO 2 produces 1 mol of chlorine. Therefore, the number of moles of chlorine produced from 0 . 35 mol of MnO 2 = 0 . 35 mol . We know that, 1 mol of Cl 2 = 22 . 7 dm 3 at 273 . 15 K and 100 kPa pressure. Therefore, 0 . 35 mol Cl 2 = 0 . 35 × 22 . 7 = 7 . 945 dm 3 . Hence, the theoretical yield of chlorine is 7 . 945 dm 3 .

Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions

Author:Sergey Bylikin, Gary Horner & Brian Murphy

Sergey Bylikin Chemistry Solutions for Exercise - Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions

Attempt the practice questions on Chapter 1: Stoichiometric Relationships, Exercise 10: Quick questions with hints and solutions to strengthen your understanding. Oxford IB Diploma Programme Chemistry Course Companion solutions are prepared by Experienced Embibe Experts.

Questions from Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions with Hints & Solutions

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 1

Butane lighters work by the release and combustion of pressurised butane:

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \overset{}{\to} 8 {CO}_{2} (g) + 10 H_{2} O (l)$

Determine the limiting reagent in the reaction, $20$ molecules of $C_{4} H_{10}$ and $100$ molecules of $O_{2}$ .

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 1

Butane lighters work by the release and combustion of pressurised butane:

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \overset{}{\to} 8 {CO}_{2} (g) + 10 H_{2} O (l)$

Determine the limiting reagent in the reaction, $10$ molecules of $C_{4} H_{10}$ and $91$ molecules of $O_{2}$ .

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 1

Butane lighters work by the release and combustion of pressurised butane:

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \overset{}{\to} 8 {CO}_{2} (g) + 10 H_{2} O (l)$

Determine the limiting reagent in the reaction, $0.20 mol$ of $C_{4} H_{10}$ and $2.6 mol$ of $O_{2}$ .

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 1

Butane lighters work by the release and combustion of pressurised butane:

$2 C_{4} H_{10} (g) + 13 O_{2} (g) \overset{}{\to} 8 {CO}_{2} (g) + 10 H_{2} O (l)$

Determine the limiting reagent in the reaction, $8.72 g$ of $C_{4} H_{10}$ and $28.8 g$ of $O_{2}$ .

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 2

Two aqueous solutions, one containing $5.3 g$ of sodium carbonate and the other $7.0 g$ of calcium chloride, are mixed together. A precipitation reaction occurs:

${Na}_{2} {CO}_{3} (aq) + {CaCl}_{2} (aq) \overset{}{\to} {CaCO}_{3} (s) + 2 NaCl (aq)$

Determine the limiting reagent and the mass, in $g$ , of precipitate formed.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 5

Chlorine gas is produced by the reaction of hydrochloric acid, and the oxidising agent manganese( $IV$ ) oxide, ${MnO}_{2}$ :

${MnO}_{2} (s) + 4 HCl (aq) \overset{}{\to} {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O (l)$

At $273.15 K$ and $100 kPa$ , $58.34 g$ of $HCl$ reacts with $0.35 mol$ of ${MnO}_{2}$ to produce $7.056 {dm}^{3}$ of chlorine gas. Deduce the limiting reagent.

EASY

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 1 - Stoichiometric Relationships>Quick questions>Q 5

Chlorine gas is produced by the reaction of hydrochloric acid, and the oxidising agent manganese( $IV$ ) oxide, ${MnO}_{2}$ :

${MnO}_{2} (s) + 4 HCl (aq) \overset{}{\to} {MnCl}_{2} (aq) + {Cl}_{2} (g) + 2 H_{2} O (l)$

At $273.15 K$ and $100 kPa$ , $58.34 g$ of $HCl$ reacts with $0.35 mol$ of ${MnO}_{2}$ to produce $7.056 {dm}^{3}$ of chlorine gas. Calculate the theoretical yield of chlorine.

Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions

Sergey Bylikin Chemistry Solutions for Exercise - Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions

Questions from Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Stoichiometric Relationships, Exercise 10: Quick questions with Hints & Solutions

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