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Viktor [21]
3 years ago
5

Given the following chemical reaction: 2 O2 + CH4 CO2 + 2 H2O What mass of CH4 is required to completely react with 100 grams of

O2?
Chemistry
2 answers:
Amanda [17]3 years ago
5 0
2 O2 + CH4 CO2 + 2 H2O 

<span>What mass of CH4 is required to completely react with 100 grams of O2? </span>

<span>mass CH4 = 100 g O2 x (1 mol O2 / 32 g) x (1 mol CH4 / 2 mol O2) x (16.05 g / 1 mol CH4) </span>
<span> 25 grams CH4 


</span>
sineoko [7]3 years ago
3 0

Answer : The mass methane required is, 25 g

Explanation : Given,

Mass of oxygen gas = 100 g

Molar mass of oxygen gas = 32 g/mole

Molar mass of methane gas = 16 g/mole

First we have to calculate the moles of O_2.

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{100g}{32g/mole}=3.125moles

Now we have to calculate the moles of CH_4.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 moles of O_2 react with 1 mole of CH_4

So, 3.125 moles of O_2 react with \frac{3.125}{2}=1.563 moles of CH_4

Now we have to calculate the mass of CH_4

\text{Mass of }CH_4=\text{Moles of }CH_4\times \text{Molar mass of }CH_4

\text{Mass of }CH_4=(1.563mole)\times (16g/mole)=25.008g=25g

Therefore, the mass methane required is, 25 g

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Answer:

Zn(s) + Fe(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Fe(s)

Explanation:

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Zn(s) + Fe(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Fe(s)

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3 0
3 years ago
Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
kati45 [8]

<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

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