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KatRina [158]
9 months ago
5

Calculate the AHrxn from the AH of formation for the following reaction. C2H4(g) + 302(g) 2C02(g) + 2H20(1). Formation AH values

for C2H4(g) = 52.30 kJ/mol, for 02(g) = 0 kJ/mol, for CO2(g) = -393.5 kJ/mol and for H20(1) =-285.8kJ/mol.A. -1305kJB. 1350kJC. 1411kJD. -1411kJ
Chemistry
1 answer:
RSB [31]9 months ago
4 0

Answer

A. -1305 kJ

Explanation

Given:

Equation: C2H4(g) + 302(g) ---- > 2C02(g) + 2H20(l).

Formation ΔH values:

​​for C2H4(g) = 52.30 kJ/mol,

for 02(g) = 0 kJ/mol,

for CO2(g) = -393.5 kJ/mol, and

for H20(1) = -285.8kJ/mol.

What to find:

The ΔHrxn from the ΔH of formation for the given reaction.

Step-by-step solution:

ΔHrxn = (Sum of ΔH formation for the product) - (Sum of ΔH formation for the reactants).

ΔHrxn = (ΔH for 2CO2(g) + ΔH for 2H2O(l)) + (ΔH for C2H4(g) + ΔH for 3O2(g))

ΔHrxn = [(2 x -393.5) + (2 x -285.8)] + [52.30 + (3 x 0)]

ΔHrxn =(-787.0 - 571.6) + (52.30 + 0)

ΔHrxn = -1358.6 + 52.30

ΔHrxn = -1306.3 kJ

so the closest answer is A. -1305 kJ

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How many moles of oxygen are formed when 58.6 g of KNO3 decomposes according to the following reaction? 4 KNO3(s) → 2 K2O(s) + 2
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Explanation:

Moles are calculated as the given mass divided by the molecular mass.

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moles = ( mass / molecular mass )

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moles of KNO₃ = 58.6 g / 101 g / mol

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