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What volume (in L) of H2 is required to produce NH3 if we have 22.4L of N2 available at STP and no H2 is left ?

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\\\text { Simply, the reaction is: } N_{2}+3 H_{2}=2 N H_{3} \\ \text{22.4 L of N}_{2} \text{ gas contains 1 mole of the gas N}_{2} \\ \text{According to equation, we need 3 moles of H}_{2} \text{ gas for 1 mole of N}_{2}. \\\\ \text{3 moles of H}_{2} = 3 \times 22.4 L = 67.2 L

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