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timofeeve [1]
3 years ago
10

Calculate ΔH when 2 moles of ethanol (C2H5OH) reacts with excess oxygen according to the following thermochemical equation?C2H5O

H + 3O2 → 2CO2 + 3H2O ΔH = −1366.7 kJ
Chemistry
1 answer:
klio [65]3 years ago
5 0

Answer:

–2733.4 KJ

Explanation:

The balanced equation for the reaction is given below:

C₂H₅OH + 3O₂ —> 2CO₂ + 3H₂O

ΔH = −1366.7 kJ

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Finally, we shall determine the enthalpy change (ΔH) produced by the reaction of 2 moles of C₂H₅OH. This can be obtained as follow:

From the balanced equation above,

1 mole of C₂H₅OH reacted to produce enthalpy change (ΔH) of −1366.7 kJ.

Therefore, 2 moles of C₂H₅OH will react to produce enthalpy change (ΔH) of = 2 × −1366.7 = –2733.4 KJ.

Thus, enthalpy change (ΔH) obtained is –2733.4 KJ

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How many grams of NO can be produced if 204 g of NO2 is mixed with 58.1 g of H2O?
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44.4 grams of NO can be produced

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Step 1: Data given

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Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 204 grams / 46.0 g/mol

Moles NO2 = 4.43 moles

Step 4: Calculate moles H2O

Moles H2O = 58.1 grams / 18.02 g/mol

Moles H2O = 3.22 moles

Step 5: Calculate limiting reactant

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

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Step 7: Calculate mass NO

Mass NO = 1.48 moles * 30.01 g/mol

Mass NO = 44.4 grams

44.4 grams of NO can be produced

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3 years ago
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