of reacts with of to form . If is the limiting reagent and of is formed, then: — Mole Concept and Some Basic Concepts of Chemistry Chemistry Question

💡 Solution & Explanation

$N_2 + 3H_2 \rightarrow 2NH_3$. <br>$2 \text{ moles } NH_3 (34 \text{ g})$ requires $3 \text{ moles } H_2 (6 \text{ g})$. <br>If $H_2$ is limiting, we must have used exactly 6 g of $H_2$, but started with more $N_2$. <br>If $H_2$ is limiting, start $Y=6$ and $X > 28$. If $Y < 6$, yield would be lower. So $Y$ must be 6.