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A sample of octane undergoes combustion according to the equation 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O ΔH°rxn = -11018 kJ. What mas

Anna71 [15]

Answer:

$\large \boxed{\text{528.7 g} }$

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32.00

2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol: 7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

$\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}$

2. Mass of O₂

$\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}$

3 0

3 years ago

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

4 years ago

Adding energy to a solid bar of gold may result in which of the following outcomes

____ [38]

An increase in motion and less attraction between particles

4 0

3 years ago

Can someone help me with 3

lozanna [386]

Solidification i believe

5 0

4 years ago

What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociatio

evablogger [386]

Answer:

1.3×10⁻³ M

Explanation:

Hello,

In this case, given the dissociation reaction of acetic acid:

$CH_3CO_2H(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3CO_2^-(aq)$

We can write the law of mass action for it:

$Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}$

Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change $x$ due to the dissociation extent, we are able to rewrite it as shown below:

$1.8x10^{-5}=\frac{x*x}{0.100-x}$

Thus, via the quadratic equation or solve, we obtain the following solutions:

$x_1=-0.00135M\\x_2=0.00133M$

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.

Regards.

8 0

4 years ago

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