Question

part a. stoichiometry
given the reaction:
2 c₈h₁₈ + 25 o₂ → 16 co₂ + 18 h₂o
25.00 g of c₈h₁₈ are combusted. how many grams of oxygen are used? how many grams of co₂ are produced?
nitroglycerin detonates, according to the reaction:
4 c₃h₅n₃o₉ → 12 co₂ + 6 n₂ + o₂ + 10 h₂o
2.50 kg of nitroglycerin detonate. how many kg of co₂ and h₂o are produced?
consider the reaction:
4 nh₃ + 5 o₂ → 4 no + 6 h₂o
10.0 g of nh₃ react with 25.0 g of o₂. what is the limiting reactant?

Explanation:

Part 1: Calculate grams of oxygen used and grams of CO₂ produced for the combustion of C₈H₁₈

Step1: Calculate molar mass of C₈H₁₈

The molar mass of C₈H₁₈ ($M_{C_8H_{18}}$) is $8\times12.01 + 18\times1.01=114.23$ g/mol.

Step2: Calculate moles of C₈H₁₈

The number of moles of C₈H₁₈ ($n_{C_8H_{18}}$) is $n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{25.00}{114.23}\approx0.219$ mol.

Step3: Determine moles of O₂ and CO₂ using stoichiometry

From the reaction $2C_8H_{18}+25O_2
ightarrow16CO_2 + 18H_2O$, the mole - ratio of C₈H₁₈ to O₂ is 2:25 and to CO₂ is 2:16.
Moles of O₂ ($n_{O_2}$) used: $n_{O_2}=\frac{25}{2}\times n_{C_8H_{18}}=\frac{25}{2}\times0.219 = 2.7375$ mol.
Molar mass of O₂ is $M_{O_2}=2\times16 = 32$ g/mol. Mass of O₂ ($m_{O_2}$) used is $m_{O_2}=n_{O_2}\times M_{O_2}=2.7375\times32 = 87.6$ g.
Moles of CO₂ ($n_{CO_2}$) produced: $n_{CO_2}=\frac{16}{2}\times n_{C_8H_{18}}=8\times0.219 = 1.752$ mol.
Molar mass of CO₂ is $M_{CO_2}=12.01+2\times16 = 44.01$ g/mol. Mass of CO₂ ($m_{CO_2}$) produced is $m_{CO_2}=n_{CO_2}\times M_{CO_2}=1.752\times44.01\approx77.1$ g.

Part 2: Calculate kg of CO₂ and H₂O produced from the detonation of nitroglycerin

Step1: Calculate molar mass of C₃H₅N₃O₉

The molar mass of C₃H₅N₃O₉ ($M_{C_3H_5N_3O_9}$) is $3\times12.01+5\times1.01 + 3\times14.01+9\times16=227.09$ g/mol.

Step2: Calculate moles of C₃H₅N₃O₉

The number of moles of C₃H₅N₃O₉ ($n_{C_3H_5N_3O_9}$) is $n_{C_3H_5N_3O_9}=\frac{m_{C_3H_5N_3O_9}}{M_{C_3H_5N_3O_9}}=\frac{2500}{227.09}\approx11.01$ mol.

Step3: Determine moles of CO₂ and H₂O using stoichiometry

From the reaction $4C_3H_5N_3O_9
ightarrow12CO_2+6N_2 + O_2+10H_2O$, the mole - ratio of C₃H₅N₃O₉ to CO₂ is 4:12 and to H₂O is 4:10.
Moles of CO₂ ($n_{CO_2}$) produced: $n_{CO_2}=\frac{12}{4}\times n_{C_3H_5N_3O_9}=3\times11.01 = 33.03$ mol.
Molar mass of CO₂ is $M_{CO_2}=44.01$ g/mol. Mass of CO₂ ($m_{CO_2}$) produced is $m_{CO_2}=n_{CO_2}\times M_{CO_2}=33.03\times44.01\approx1453.65$ g or 1.45 kg.
Moles of H₂O ($n_{H_2O}$) produced: $n_{H_2O}=\frac{10}{4}\times n_{C_3H_5N_3O_9}=2.5\times11.01 = 27.525$ mol.
Molar mass of H₂O is $M_{H_2O}=18.02$ g/mol. Mass of H₂O ($m_{H_2O}$) produced is $m_{H_2O}=n_{H_2O}\times M_{H_2O}=27.525\times18.02\approx496.0$ g or 0.496 kg.

Part 3: Determine the limiting reactant for the reaction of NH₃ and O₂

Step1: Calculate molar mass of NH₃ and O₂

The molar mass of NH₃ ($M_{NH_3}$) is $14.01+3\times1.01 = 17.04$ g/mol. The molar mass of O₂ ($M_{O_2}$) is $2\times16 = 32$ g/mol.

Step2: Calculate moles of NH₃ and O₂

Moles of NH₃ ($n_{NH_3}$) is $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{10.0}{17.04}\approx0.587$ mol.
Moles of O₂ ($n_{O_2}$) is $n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{25.0}{32}=0.78125$ mol.

Step3: Determine the limiting reactant using stoichiometry

From the reaction $4NH_3 + 5O_2
ightarrow4NO+6H_2O$, the mole - ratio of NH₃ to O₂ is 4:5.
For 0.587 mol of NH₃, the moles of O₂ required ($n_{O_2 - required}$) is $n_{O_2 - required}=\frac{5}{4}\times n_{NH_3}=\frac{5}{4}\times0.587 = 0.73375$ mol.
Since $0.73375\lt0.78125$ (the amount of O₂ available), NH₃ is the limiting reactant.

Answer:

• Grams of O₂ used in C₈H₁₈ combustion: 87.6 g
• Grams of CO₂ produced in C₈H₁₈ combustion: 77.1 g
• kg of CO₂ produced in nitroglycerin detonation: 1.45 kg
• kg of H₂O produced in nitroglycerin detonation: 0.496 kg
• Limiting reactant in NH₃ and O₂ reaction: NH₃