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What is the amount of mol of water produced if

9.05 x 10 to the 23rd molecules of nitrogen monoxide, NO, is
reacted?
Rxn B: 4NH3 + 6NO → 5N₂ + 6H₂O

User Aleroy
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1 Answer

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The number of moles of water produced can be calculated by first finding the number of moles of NO that are reacted.

Since 9.05 x 10^23 molecules of NO are present, the number of moles can be calculated as follows:

9.05 x 10^23 molecules NO * (1 mole NO / 6.02 x 10^23 molecules NO) = 1.50 moles NO

Next, we need to use the stoichiometry of the reaction to find the amount of water produced for the given number of moles of NO. From the equation, for every 6 moles of NO, 6 moles of water are produced. So, for 1.50 moles of NO, 1.50 moles NO / 6 NO/6 H2O = 0.25 moles of water are produced.

User Alexey  Pavlov
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