How many liters of NH3 are needed to react completely with 30.0 L of NO (at STP)?4NH3 (g)+ 6NO(g) 5N2(g)+ 6H2o(g)
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First, we need to determine the balanced chemical equation:
4NH3 (g) + 6NO (g) → 5N2 (g) + 6H2O (g)
From this equation, we can see that for every 6 moles of NO, we need 4 moles of NH3 to react completely.
Next, we need to use the ideal gas law to relate the volume of NO at STP to the moles of NO:
PV = nRT
(1 atm) (30.0 L) = n (0.0821 L•atm/mol•K) (273 K)
n = 1.23 mol NO
Using the mole ratio from the balanced equation, we can determine how many moles of NH3 are needed:
4 mol NH3 / 6 mol NO = 0.67 mol NH3
Finally, we use the ideal gas law again to relate the moles of NH3 to the volume of NH3 at STP:
PV = nRT
V = nRT/P = (0.67 mol) (0.0821 L•atm/mol•K) (273 K) / (1 atm) = 12.6 L NH3
Therefore, 12.6 L of NH3 are needed to react completely with 30.0 L of NO at STP.