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56.0 L of nitrogen gas is mixed with excess hydrogen gas and it is found that 20 L of ammonia gas is produced. The volume of unused nitrogen gas is found to be _________ L.
N2( g)+3 H2( g) → 2 NH3( g)\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})N2( g)+3 H2( g) → 2 NH3( g)
Since H2\mathrm{H}_{2}H2 is in excess and 20 L20 \mathrm{~L}20 L of ammonia gas is produced.
Hence, 2 moles NH3 ≡ 1\mathrm{NH}_{3} \equiv 1NH3 ≡ 1 mole N2 (v ∝ n)\mathrm{N}_{2} \quad(v \propto \mathrm{n})N2 (v ∝ n)
20 L NH3 ≡ 10 L N2 20 \mathrm{~L} \mathrm{NH}_{3} \equiv 10 \mathrm{~L} \mathrm{~N}_{2} 20 L NH3 ≡ 10 L N2
Volume of N2\mathrm{N}_{2}N2 left =56−10=56-10=56−10
=46 L =46 \mathrm{~L} =46 L
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