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This system has an equilibrium constant of 0.105 at 472°C: N2(g) + 3H2(g) ↔ 2NH3(g)
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<u>Answer:</u> The moles of water produced are 1.54 moles.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of ethane = 15.42 g
Molar mass of ethane = 30.07 g/mol
Putting values in above equation, we get:
The chemical equation for the combustion of ethane follows:
By Stoichiometry of the reaction:
2 moles of ethane produces 6 moles of water
So, 0.513 moles of ethane will produce = of water
Hence, the moles of water produced are 1.54 moles.
