Answer:
34.7 L of hydrogen are needed to generate 12 L of NH3 at STP.
Step-by-step explanation:
First, we need to determine the number of moles of ammonia produced by reacting 12 L of NH3 at STP (standard temperature and pressure, which is 0°C and 1 atm). At STP, one mole of any ideal gas occupies 22.4 L, so:
12 L NH3 * (1 mol NH3 / 22.4 L NH3) = 0.536 mol NH3
According to the balanced chemical equation, 3 moles of hydrogen react with 1 mole of ammonia to produce 2 moles of ammonia. Therefore, we need 1.5 times as many moles of hydrogen as ammonia, or:
0.536 mol NH3 * (3 mol H2 / 1 mol NH3) = 1.608 mol H2
Now we can use the ideal gas law, which relates the volume, pressure, temperature, and number of moles of a gas:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, the pressure is 1 atm and the temperature is 273 K. Rearranging the ideal gas law to solve for V and plugging in the values, we get:
V = nRT/P = (1.608 mol) * (0.0821 L·atm/(mol·K)) * (273 K) / (1 atm) = 34.7 L H2
34.7 L of hydrogen are needed to generate 12 L of NH3 at STP.