Chemistry, 07.03.2020 04:28 pinapunapula

When methane (CH4) burns, it reacts with oxygen gas to produce carbon dioxide and water. The unbalanced equation for this reaction iCH4(g)+O2(g)→CO2(g)+H2O(g)This type of reaction is referred to as a complete combustion reaction. A)What mass of carbon dioxide is produced from the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units. B)What mass of water is produced from the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units. C)What mass of oxygen is needed for the complete combustion of 1.10×10−3 g of methane?Express your answer with the appropriate units.

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Answer from: Eylul30

For A: The mass of carbon dioxide produced is 0.00303 grams.

For B: The mass of water produced is 0.0025 grams.

For C: The mass of oxygen gas produced is 0.0044 grams.

Explanation:

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

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

For methane:

Given mass of methane = $1.10\times 10^{-3}g$

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{1.10\times 10^{-3}g}{16g/mol}=6.875\times 10^{-5}mol$

The chemical equation for the combustion of methane follows:

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

For A:

By Stoichiometry of the reaction:

1 mole of methane produces 1 mole of carbon dioxide

So, $6.875\times 10^{-5}mol$ of methane will produce = $\frac{1}{1}\times 6.875\times 10^{-5}=6.875\times 10^{-5}mol$ of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = $6.875\times 10^{-5}$ moles

Putting values in equation 1, we get:

$6.875\times 10^{-5}mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(6.875\times 10^{-5}mol\times 44g/mol)=0.00303g$

Hence, the mass of carbon dioxide produced is 0.00303 grams.

For B:

By Stoichiometry of the reaction:

1 mole of methane produces 2 moles of water

So, $6.875\times 10^{-5}mol$ of methane will produce = $\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol$ of water

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

Molar mass of water = 18 g/mol

Moles of water = $1.375\times 10^{-4}$ moles

Putting values in equation 1, we get:

$1.375\times 10^{-4}mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.375\times 10^{-4}mol\times 18g/mol)=0.0025g$

Hence, the mass of water produced is 0.0025 grams.

For C:

By Stoichiometry of the reaction:

1 mole of methane reacts with 2 moles of oxygen gas

So, $6.875\times 10^{-5}mol$ of methane will react with = $\frac{2}{1}\times 6.875\times 10^{-5}=1.375\times 10^{-4}mol$ of oxygen gas

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

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = $1.375\times 10^{-4}$ moles

Putting values in equation 1, we get:

$1.375\times 10^{-4}mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(1.375\times 10^{-4}mol\times 32g/mol)=0.0044g$

Hence, the mass of oxygen gas produced is 0.0044 grams.

Answer from: Quest

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh $pcl_{3}$ +δh $cl_{2}$ ] - [δh $pcl_{5}$ ]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

Answer from: Quest

Yes i think so not 100 percent sure though

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