Question

use the references to access important values if needed for this question. for the following reaction, 0.243 moles of nitrogen gas are mixed with 0.497 moles of oxygen gas. nitrogen(g) + oxygen(g) → nitrogen monoxide(g) what is the formula for the limiting reactant? what is the maximum amount of nitrogen monoxide that can be produced? amount = mol

Explanation:

Step1: Balance the chemical equation

$N_2(g)+O_2(g)
ightarrow 2NO(g)$

Step2: Determine the mole - ratio of reactants

The mole - ratio of $N_2$ to $O_2$ is 1:1 from the balanced equation.

Step3: Calculate the amount of product formed from each reactant

For nitrogen gas ($N_2$), if all 0.243 moles of $N_2$ react, the moles of $NO$ produced is $2\times0.243 = 0.486$ moles (using the mole - ratio of $N_2$ to $NO$ which is 1:2).
For oxygen gas ($O_2$), if all 0.497 moles of $O_2$ react, the moles of $NO$ produced is $2\times0.497=0.994$ moles (using the mole - ratio of $O_2$ to $NO$ which is 1:2).

Step4: Identify the limiting reactant

Since $N_2$ produces less $NO$ (0.486 moles) compared to what $O_2$ would produce if it all reacted, $N_2$ is the limiting reactant.

Step5: Determine the maximum amount of product

The maximum amount of $NO$ produced is determined by the limiting reactant. So the maximum amount of $NO$ is 0.486 moles.

Answer:

The formula for the limiting reactant: $N_2$
The maximum amount of nitrogen monoxide that can be produced: 0.486 mol