$$56.0 \text{ L}$$ of nitrogen gas is mixed with excess of hydrogen gas and it is found that $$20 \text{ L}$$ of ammonia gas is produced. The volume of unused nitrogen gas is found to be ______ L.

We are given that 56.0 L of nitrogen gas is mixed with excess hydrogen gas and 20 L of ammonia is produced. We need to find the volume of unused nitrogen gas.

Write the balanced equation

$$N_2 + 3H_2 \rightarrow 2NH_3$$

Use the volume ratio (at same T and P, volume ratio = mole ratio)

From the equation: 1 volume of $$N_2$$ produces 2 volumes of $$NH_3$$.

Calculate the volume of $$N_2$$ used

Volume of $$NH_3$$ produced = 20 L

$$\text{Volume of } N_2 \text{ used} = \frac{20}{2} = 10 \text{ L}$$

Calculate unused $$N_2$$

$$\text{Unused } N_2 = 56.0 - 10 = 46 \text{ L}$$

The answer is 46.