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marysya [2.9K]
3 years ago
9

Calculate the enthalpy change for the thermite reaction: 2al(s)+fe2o3(s)→2fe(s)+al2o3(s), δh∘rxn=−850 kj when 4.00 mol of al und

ergoes the reaction with a stoichiometrically equivalent amount of fe2o3.
Chemistry
1 answer:
kolezko [41]3 years ago
6 0

<u>Given:</u>

Enthalpy change ΔH for the thermite reaction = -850 kJ

Moles of Al involved = 4

<u>To determine:</u>

Reaction enthalpy when 4 moles of Al reacts

<u>Explanation:</u>

The thermite reaction is-

2Al(s) + Fe2O3(s) → 2Fe(s) + Al2O3(s)

When 2 moles of Al react the enthalpy change is -850 kJ

therefore, when 4 moles of Al reacts, the change in enthalpy is-

= 4 moles * (-850) kJ/2 moles = -1700 kJ

Ans: Enthalpy change is -1700 kJ


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Answer:

We need 226 grams of FeS

Explanation:

Step 1: Data given

Mass of FeCl2 = 326 grams

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Step 2: The balanced equation

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Step 3: Calculate moles FeCl2

Moles FeCl2 = 326 grams / 126.75 grams

Moles FeCl2 = 2.57 moles

Step 4: Calculate moles FeS needed

For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl

For 2.57 moles FeCl2 we need 2.57 moles FeS

Step 5: Calculate mass FeS

Mass FeS = 2.57 moles * 87.92 g/mol

Mass FeS = 226 grams FeS

We need 226 grams of FeS

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Is the sweetened iced tea an electrolyte mixture?
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Yes it is

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A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of
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Answer:

a) T_{2} = 360.955\,K, P_{2} = 138569.171\,Pa\,(1.386\,bar), b) T_{2} =  347.348\,K, V_{2} = 0.14\,m^{3}

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a) The ideal gas is experimenting an isocoric process and the following relationship is used:

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Final temperature is cleared from this expression:

Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}

The number of moles of the ideal gas is:

n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}

n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}

n = 5.541\,mol

The final temperature is:

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}

T_{2} = 360.955\,K

The final pressure is:

P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}

P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)

P_{2} = 138569.171\,Pa\,(1.386\,bar)

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}

Final temperature is cleared from this expression:

Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})

T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}

T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}

T_{2} =  347.348\,K

The final volume is:

V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}

V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})

V_{2} = 0.14\,m^{3}

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Kilo pascal is denoted as "KPa"

Millimeter mercury is denoted as " mmHg"

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