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3 years ago

What is the product of the unbalanced combustion reaction below? C4H10(g) + O2(g) →

Chemistry

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

Option C . CO2(g) + H2O(g)

Explanation:

When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

Thus, the product of the unbalanced combustion reaction is:

CO2(g) + H2O(g)

Thus, we can balance the equation as follow:

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)

There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:

C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)

Thus, the equation is balanced.

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You have 4 moles of oxygen gas in a flask. 4 moles of helium gas is added. What happens to the total pressure of the gases in th

ASHA 777 [7]

Answer: The correct option is (c). The total pressure doubles.

Solution:

Initially, only 4 moles of oxygen gas were present in the flask.

$p_{O_2}=Tp_1\times X_{O_2}$ ( $X_{O_2}=\frac{4}{4}$ ) ( according to Dalton's law of partial pressure)

$p_{O_2}=Tp_1\times 1=Tp_1$ ....(1)

$Tp_1$ = Total pressure when only oxygen gas was present.

Final total pressure when 4 moles of helium gas were added:

$X'_{O_2}=\frac{4}{8}=\farc{1}{2},X_{He}=\frac{4}{8}=\frac{1}{2}$

partial pressure of oxygen in the mixture :

Since, the number of moles of oxygen remains the same, the partial pressure of oxygen will also remain the same in the mixture.

$p_{O_2}=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

$Tp_2$ = Total pressure of the mixture.

from (1)

$Tp_1=Tp_2\times X'_{O_2}=Tp_2\times \frac{1}{2}$

On rearranging, we get:

$Tp_2=2\times Tp_1$

The new total pressure will be twice of initial total pressure.

7 0

3 years ago

Which has the highest electronegativity value?

AnnZ [28]

Which has the highest electronegativity value?
A
hydrogen

B
calcium

C
helium

D
fluorine d because fluorine has a higher group number

8 0

3 years ago

A new substance is formed when the?

quester [9]

Answer:

A chemical reaction happens when substances break apart or combine to form one or more new substances.

Explanation:

hope its right.

4 0

3 years ago

The combustion reaction of isopropyl alcohol is given below: C 3 H 7 O H ( l ) + 9 2 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) T

jek_recluse [69]

Answer:

the heat of formation of isopropyl alcohol is -317.82 kJ/mol

Explanation:

The heat of combustion of isopropyl alcohol is given as follows;

C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)

The heat of combustion of CO₂ and H₂O are given as follows

C (s) + O₂ (g) → CO₂(g) = −393.50 kJ

H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ

Therefore we have

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as

3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =

4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol

-1180.5 - 1143.32 +2006 = -317.82 kJ/mol

Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.

3 0

3 years ago

At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

wlad13 [49]

Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

[Cl₂O] = 0.02351 mol/liter

Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

Thus the concentrations are:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

3 0

3 years ago