If a mixture containing 0.50 moles of N2 and 1.8 moles of H2 is allowed to react according to the equation N2 3H2 2NH3 Group of
1 answer:
Answer: hydrogen is the limiting reactant.
Explanation:
We have the equation .
This means that for every mole of nitrogen consumed, 3 moles of hydrogen are consumed.
- Considering the nitrogen, the reaction can occur 0.50 times.
- Considering the hydrogen, the reaction can occur 1.8/3 = 0.6 times.
Therefore, <u>hydrogen</u> is the limiting reactant.
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The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.
The reaction is:
2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O (1)
The enthalpy of reaction (1) is given by:
(2)
Where:
r: is for reactants
p: is for products
The bonds of the compounds of reaction (1) are:
- 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
- 7O₂: 7 moles of 1 O=O bond
- 4CO₂: 4 moles of 2 C=O bonds
- 6H₂O: 6 moles of 2 H-O bonds
Hence, the enthalpy of reaction (1) is (eq 2):
Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.
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