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For the following equation: 4NH3(g) + 7O2(g) ® 4NO2(g) + 6H2O(g), how many moles of ammonia (NH3) will be required to produce 10.0 mol of water?
a.) 4.00 mol
b.) 10.0 mol
c.) 6.67 mol
d.) 5.00 mol

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Answer:

6.67 moles of NH₃

Explanation:

From the equation of reaction,

4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g),

4 moles of NH₃ will produce 6 moles of 6H₂O,

X moles of NH₃ will produce 10 moles of H₂O

X = (10 * 4) / 6

X = 6.67 moles of NH₃

6.67 moles of NH₃ will produce 10 moles of H₂O

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