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Slav-nsk [51]
2 years ago
7

Using table 9.4, calculate an approximate enthalpy (in kj) for the reaction of 1.02 g gaseous methanol (ch3oh) in excess molecul

ar oxygen to form gaseous carbon dioxide and gaseous water. (hint, remember to first write the balanced equation.)
Chemistry
1 answer:
Tanya [424]2 years ago
6 0

<u>Given:</u>

Mass of methanol = 1.02 g

<u>To determine:</u>

Enthalpy for the reaction of 1.02 g of methanol with excess O2

<u>Explanation:</u>

Balanced equation-

2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)

The reaction enthalpy is given as:

ΔHrxn = ∑nH°f(products) - ∑nH°f(reactants)

where n = number of moles

H°f = standard enthalpy of formation.

ΔHrxn = [2H°f(CO2(g)) + 4H°f(H2O(g))] - [2H°f(CH3OH(g)) + 3H°f(O2(g))]

           = [2(-393.5) + 4(-241.8)]-[2(-201.5) + 3(0)] = -1351.2 kJ

Now, 1 mole of CH3OH = 32 g

The calculated ΔHrxn corresponds to 2 moles of CH3OH. i.e.

The enthalpy change for 64 g of Ch3OH = -1351.2 kJ

Therefore, for 1.02g gaseous methanol we have:

ΔH = 1.02 * -1351.2/64 = -21.5 kJ

Ans: The enthalpy for the given reaction is -21.5 kJ



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please find the complete question in the attached file.

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No of moles in n_{CH_3CO_2H} = 77 \ g \times  \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}  

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Mass of water (H_2O)= 42 \ g

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= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O

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X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\

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During the distillation nitrogen gas is obtained first, then argon and oxygen. What can u say about the boiling points of these
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A 17.11 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 21.71 g CO2 and 5.
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Answer:  The empirical formula and the molecular formula of the organic compound is CHO and C_4H_4O_4 respectively.

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2 = 21.71 g

Mass of H_2O= 5.926 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 21.71 g of carbon dioxide, =\frac{12}{44}\times 21.71=5.921g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 5.926 g of water, =\frac{2}{18}\times 5.926=0.658g of hydrogen will be contained.

Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53  g

Mass of C = 5.921 g

Mass of H = 0.658 g

Mass of O = 10.53 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.921g}{12g/mole}=0.493moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.658g}{1g/mole}=0.658moles

Mass of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{10.53g}{16g/mole}=0.658moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.493}{0.493}=1

For H =\frac{0.658}{0.493}=1

For O=\frac{0.658}{0.493}=1

The ratio of C : H: O =  1: 1: 1

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n=\frac{\text {Molecular mass}}{\text {Equivalent mass}}=\frac{104.1}{29}=4

Thus molecular formula = n\times {\text {Empirical formula}}=4\times CHO=C_4H_4O_4

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