Question 6 (1 point) Hydrogen gas and nitrogen gas can react to form ammonia according to this equation: N2 (g) + m 3H2 (g) --> 2NH3(g) How many liters of nitrogen gas are required to produce 4 moles of NH3 (g)? Assume conditions of standard temperature and pressure (STP). Hint: One mole of gas occupies 22.4 liters under standard conditions of temperature and pressure (STP).

44.8 L of N₂.

Explanation:

We'll begin by writing the balanced equation for the. This is illustrated below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted to produce 2 moles of NH₃.

Therefore, Xmol of N₂ will react to produce 4 moles of NH₃ i.e

Xmol of N₂ = (1 × 4)/2

Xmol of N₂ = 2 moles

Thus, 2 moles of N₂ reacted to produce 4 moles of NH₃.

Finally, we shall determine the volume of N₂ required for the reaction. This can be obtained as follow:

1 mole of N₂ occupies 22.4 L at STP.

Therefore, 2 moles of N₂ will occupy = (2 × 22.4) = 44.8 L

Thus, 44.8 L of N₂ is needed to produce 4 moles of NH₃.

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