Chemistry, 03.07.2019 20:30 zay179

Consider the following reaction: 2ch3oh(g)  2ch4(g) + o2(g) δh = +252.8 kj a) calculate the amount of heat transferred when 24.0 g of ch3oh(g) is decomposed by this reaction at constant pressure. b) for a given sample of ch3oh, the enthalpy change during the reaction is 82.1 kj. how many grams of methane gas are produced?

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Answer from: hannahmyung1113

For a: The amount of heat transferred for the given amount of methanol is 94.6736 kJ.

For b: The mass of methane gas produced will be 10.384 g.

Explanation:

For the given chemical reaction:

$2CH_3OH(g)\rightarrow 2CH_4(g)+O_2(g);\Delta H=+252.8kJ$

For a:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of methanol = 24.0 g

Molar mass of methanol = 32.04 g/mol

Putting values in above equation, we get:

$\text{Moles of methanol}=\frac{24.0g}{32.04g/mol}=0.749mol$

By Stoichiometry of the reaction:

For every 2 moles of methanol, the amount of heat transferred is +252.8 kJ.

So, for every 0.749 moles of methanol, the amount of heat transferred will be = $\frac{252.8}{2}\times 0.749=94.6736kJ$

Hence, the amount of heat transferred for the given amount of methanol is 94.6736 kJ.

For b:

By Stoichiometry of the reaction:

252.8 kJ of energy is absorbed when 2 moles of methane gas is produced.

So, 82.1 kJ of energy will be absorbed when = $\frac{2}{252.8}\times 82.1=0.649mol$ of methane gas is produced.

Now, calculating the mass of methane gas from equation 1, we get:

Molar mass of methane gas = 16 g/mol

Moles of methane gas = 0.649 moles

Putting values in equation 1, we get:

$0.649mol=\frac{\text{Mass of methane gas}}{16g/mol}\\\\\text{Mass of methane}=10.384g$

Hence, the mass of methane gas produced will be 10.384 g.

Answer from: Quest

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation:

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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Consider the following reaction: 2ch3oh(g)  2ch4(g) + o2(g) δh = +252.8 kj a) calculate the amount...

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