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3 years ago

How much heat is released when 6.00 g of methane is combusted using the reaction below? CH4 + 2O2 ? CO2 + 2H2O ?Hrxn = -890. KJ

Chemistry

2 answers:

goblinko [34]3 years ago

8 0

Answer : The amount of heat released is, -121.04 KJ

Solution : Given,

Enthalpy of reaction, $\Delta H_{rxn}$ = -890 KJ

Mass of methane = 6 g

Molar mass of methane = 44 g/mole

First we have to calculate the moles of methane.

$\text{Moles of methane}=\frac{\text{Mass of methane}}{\text{Molar mass of methane}}=\frac{6g}{44g/mole}=0.136moles$

The given balanced reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

1 mole of methane releases heat = -890 KJ

0.136 moles of methane releases heat = $\frac{0.136moles}{1mole}\times -890KJ=-121.04KJ$

Therefore, the amount of heat released is, -121.04 KJ

anastassius [24]3 years ago

3 0

Answer:

-333.75 KJ

Explanation:

From the reaction CH₄ + 2O₂ ⇒ CO₂ + 2H₂O ΔHrxn = -890 KJ.

From the reaction, 1 mole of CH₄ combusts to give -890 KJ.

We now calculate the number of moles, n of CH₄ in 6.00g

n = mass of CH₄/molar mass of CH₄

molar mass of CH₄ = 12 g/mol + 4 × 1 g/mol = (12 + 4) g/mol = 16 g/mol

mass of CH₄ = 6.00 g

n = 6.00 g /16 g/mol = 0.375 mol

So we have 0.375 mol of CH₄ in 6.00g

From the reaction, 1 mole of CH₄ produces -890 KJ

then 0.375 mol produces 0.375 mol × -890 KJ/1 mol = -333.75 KJ

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8 0

2 years ago

Read 2 more answers

Help please with question number 16)

Zanzabum

Answer:

Mg²⁵ = 10.00%

Mg²⁶ = 45.04%

Mg²⁴ = 44.96%

Explanation:

Given data:

Atomic mass of Mg²⁶ = 25.983

Atomic mass of Mg²⁵ = 24.986

Atomic mass of Mg²⁴ = 23.985

Abundance of Mg²⁵ = 10.00%

Abundance of Mg²⁶ = ?

Abundance of Mg²⁴ = ?

Solution:

Average atomic weight of Mg = 25.983 + 24.986+ 23.985 / 3

Average atomic weight of Mg = 74.954/3

Average atomic weight of Mg = 24.985 amu

Abundance of

Mg²⁵ = 10.00

Mg²⁶ = x

Mg²⁴ = 100- 10 - x = 90 -x

Formula:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100

24.985 = (0.1×24.986)+(90-x×23.985) + ( x ×25.983 ) /100

24.985 = 249.86 + 2158.65 - 23.985x + 25.983x / 100

24.985 = 2408.51 + 1.998 x / 100

2498.5 = 2408.51 + 1.998 x

1.998 x = 2498.5 - 2408.51

1.998 x = 89.99

x = 89.99 /1.998

x = 45.04

Now we put the value of x:

Mg²⁵ = 10.00

Mg²⁶ = x (45.04)

Mg²⁴ = 90 -x (90 - 45.04 = 44.96)

5 0

2 years ago

A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction

Molodets [167]

Answer: The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For NaOH:

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol$

For sulfuric acid:

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol$

The chemical equation for the reaction of NaOH and sulfuric acid follows:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, $2.925\times 10^{-3}$ moles of KOH will react with = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = $(4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol$

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, $2.925\times 10^{-3}$ moles of KOH will produce = $\frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol$ of sodium sulfate

For sodium sulfate:

Moles of sodium sulfate = $1.462\times 10^{-3}moles$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M$

For sulfuric acid:

Moles of excess sulfuric acid = $3.498\times 10^{-3}mol$

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M$

For NaOH:

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0}{0.050}=0M$

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0

3 years ago

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tankabanditka [31]

Answer:

True

Explanation:

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8. Which of the following is TRUE about limiting and excess reactants?

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B. The limiting reactant determines the max amount of product that can be formed

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