Respuesta :
Answer:
The volume of the jar needed in order to hold enough oxygen for complete combustion is 24.2 L
(a) C₂H₅OH + 2O₂ -------> 2CO₂ + 3H₂O
(b) The moles of oxygen needed to completely combust the ethanol is 0.0429 moles
(c) The Partial pressure of oxygen is 4346.84 Pa
(d) the volume of oxygen (in L) needed in the jar is 1.038 L
Explanation:
Given;
volume of ethanol, V = 1.25 mL
density of ethanol, ρ = 0.789 g/mL
mass of ethanol, m = ρV = (1.25 mL) x (0.789 g/mL) = 0.986g
The volume of the dry air is calculated as;
[tex]P_d = \rho _d R_d T\\\\P_d = (\frac{M_d}{V_d} )R_d T\\\\V_d = (\frac{M_d}{P_d} )R_d T\\\\V_d = (\frac{0.02897 \ (kg)}{101325 \ (Pa)} )*287 \ (J/K.kg)* 295\ (K)\\\\V_d = 0.0242 \ m^3\\\\V_d = 24.2 \ L[/tex]
(a) a balanced chemical reaction for the combustion of ethanol
C₂H₅OH + 2O₂ -------> 2CO₂ + 3H₂O
(b) The moles of oxygen needed to completely combust the ethanol.
[tex]=0.986 g \ C_2H_5OH *(\frac{1 \ mol \ C_2H_5OH}{46g \ C_2H_5OH} )*(\frac{2 \ Mol \ O_2}{1 \ mol \ C_2H_5OH} )\\\\ = 0.0429 \ mols.[/tex]
(c) the partial pressure of oxygen in the jar
Total pressure of air in the jar, Pt = 101,325 Pa
Partial pressure of oxygen = nPt
= 0.0429 x 101,325 Pa = 4346.84 Pa
(d) the volume of oxygen (in L) needed in the jar
Total volume of air in the jar = 24.2 L
The fraction of this volume by oxygen = 0.0429 mols x 24.2 L = 1.038 L
Dry air is composed of many molecules and compounds. The major quantity of molecules present in the dry air is Nitrogen, followed by oxygen, argon, carbon dioxide and many more. Dry air also has some amount of water vapor in it.
The volume of the jar needed to hold enough oxygen for complete combustion can be calculated by:
- Volume of ethanol= 1.25 ml
- Density of ethanol = 0.789 g/ml
- Mass of ethanol can be calculated by:
= density × volume
= 1.25 ml x 0.789 g/ml
= 0.986 g
The volume of the dry air is calculated by the following:
[tex]\text{P}_d &= \rho_d \text{R}_d \text{T}[/tex]
[tex]\text{P}_d &= \dfrac{{\text M}_d}{\text{V}_d}} \text{R}_d \text{T}[/tex]
[tex]\text{V}_d &= \dfrac{{\text M}_d}{\text{V}_d}} \text{R}_d \text{T}[/tex]
[tex]\text{V}_d &= 0.0242\; m^{3}[/tex]
[tex]\text{V}_d &= 24.2\;\text L[/tex]
Hence, the volume of the dry air is 24.2 L.
a. Balanced chemical equation for the combustion of ethanol can be written as:
C₂H₅OH + 2 O₂ -------> 2 CO₂ + 3H₂O
One molecule of ethanol reacts with two molecules of oxygen to produce two molecules of carbon dioxide and three water molecules.
b. To calculate the moles of oxygen needed for complete combustion of ethanol can be calculated as:
[tex]\begin{aligned}&=0.986g\;\text{C}_2\text{H}_5\text{OH}\times\left(\frac{1\;\text{mol}\;\text{C}_2\text{H}_5\text{OH}}{46\;g\;\text{C}_2\text{H}_5\text{OH}} \right )\times\left(\frac{2\;\text{Mol\;O}_2}{1\;\text{mol}\text{C}_2\text{H}_5\text{OH}} \right )\\&=0.0429\text{mols} \end{aligned}[/tex]
Hence, 0.0429 moles of oxygen will be needed to completely burn ethanol.
c. The partial pressure of the oxygen in the jar can be calculated by the following formula:
Given,
- Total pressure of air in the jar (Pt) = 101,325 Pa
- The partial pressure of oxygen:
= nPt
= 0.0429 x 101,325 Pa
= 4346.84 Pa
Hence, the partial pressure of oxygen will be 4346.84 Pa.
d. The volume of oxygen (in L) needed in the jar can be calculated using the volume of the jar.
- The total volume of air in the jar = 24.2 L
- The fraction of this volume by oxygen:
= 0.0429 moles x 24.2 L
= 1.038 L
Therefore, the volume of the jar needed to hold enough oxygen for complete combustion is 24.2 L.
To learn more about combustion reaction follow the given link:
https://brainly.com/question/25092860
