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3 years ago

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A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 126 g of solutio

n. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 ∘C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water. Enter your answer in kilojoules per mole of compound to three significant figures.

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

<u>0.976 kJ/mol</u>

Explanation:

A foam coffe cup is considered a perfectly insulated system: heat energy is not exchanged with the surroundings.

Under that assumption, the heat released by the chemical reaction is equal to the heat absorbed by the system.

1. Heat absorbed by the system:

Use the equation Heat = Q = m × C × ΔT, with:

m = 126 g (the amount of solution produced)
C = specific heat of pure water = 4.186 J/gºC
ΔT = increase of temperature = 24.70 ºC - 21.00ºC = 3.70ºC

Q = 126g × 4.186J/gºC × 3.70ºC = 1,951.5J

<em><u></u></em>

<em><u>2. Enthalpy of the reaction</u></em>

The enthalpy must be reported in kJ/mol.

Then, convert juoles to kilojoules, dividing by 1,000; and divide by 2.00 moles, which is the amount of compound that reacted:

ΔHrxn = 1,951.5J × (1kJ / 1,000J) × (1 / 2mol) ≈ 0.9758 kJ/mol

Round to <em>3 significant figures</em>: 0.976 kJ/mol

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Consider a mixture of NaCl and NaNO3 that is 31.8% Na by mass. Calculate the percentage by mass of NaCl in the mixture.

lutik1710 [3]

<u>Answer:</u> The percentage by mass of NaCl in the mixture is 38.5 %

<u>Explanation:</u>

To calculate the mass percentage of element in compound, we use the equation:

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<u>For NaCl:</u>

Mass of sodium element = 23 g

Mass of NaCl = 58.5 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{58.5g}\times 100=39.32\%$

Mass fraction of sodium metal in NaCl = 0.3932

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Mass of sodium element = 23 g

Mass of $NaNO_3$ = 85 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{85g}\times 100=27.1\%$

Mass fraction of sodium metal in sodium nitrate = 0.271

Let us assume the mass fraction of NaCl in the mixture is 'x'

So, the mass fraction of $NaNO_3$ in the mixture will be '(1-x)'

We are given:

Percent by mass of Na in the mixture = 31.8 %

Mass fraction of Na in the mixture = 0.318

Evaluating the mass fraction of NaCl in the mixture:

$[(x\times 0.3932)+((1-x)\times 0.271)]=0.318$

x = 0.385

Percent by mass of NaCl in the mixture will be = $(0.385\times 100)=38.5\%$

Hence, the percentage by mass of NaCl in the mixture is 38.5 %

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Answer:

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Hello!

In this case, since the undergoing chemical reaction is:

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Best regards!

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