What is the correct number of moles of NO that is produced from 13.2 moles of oxygen gas in the presence of excess of ammonia? 4NH3 (g)  +  5O2(g)    ----------->    4NO(g)   +   6H2O(l)

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What is the correct number of moles of NO that is produced from 13.2 moles of oxygen gas in the presence of excess of ammonia?

4NH3 (g)  +  5O2(g)    ----------->    4NO(g)   +   6H2O(l)

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