If 28 g of $N_{2}$ $N_2$ and 25 g of $H_{2}$ $H_2$ are reacted together, which one would be the limiting reactant?

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If 28 g of $N_{2}$ $N_2$ and 25 g of $H_{2}$ $H_2$ are reacted together, which one would be the limiting reactant?

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Answer 1 · 2017-05-22T00:53:03

$N_{2}$ $N_2$

Explanation:

If we assume here the product is ammonia, $N H_{3}$ $NH_3$ , the equation looks like

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ $N_2(g) + 3H_2(g) rightleftharpoons 2NH_3(g)$

To find the limiting reactant, let's convert the given masses of $N_{2}$ $N_2$ and $H_{2}$ $H_2$ to moles using their molar masses, and then divide the calculated number by the coefficient in front of it in the equation. Whichever number in the end is lower, that reactant is limiting.

Using the molar masses of $N_{2}$ $N_2$ and $H_{2}$ $H_2$ :

$28cancel(gN_2)((1molN_2)/(28.02cancel(gN_2))) = color(red)(1.0 mol N_2$

$25cancel(gH_2)((1molH_2)/(2.02cancel(gH_2))) = (12molH_2)/(3"(coefficient)") = color(blue)(4.0 mol H_2$

Since $N_2$ is present in a lower relative amount than $H_2$ , nitrogen is the limiting reactant.

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If 28 g of $N_{2}$ $N_2$ and 25 g of $H_{2}$ $H_2$ are reacted together, which one would be the limiting reactant?

1 Answer

Explanation:

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If 28 g of N_2N2N_2 and 25 g of H_2H2H_2 are reacted together, which one would be the limiting reactant?

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If 28 g of $N_{2}$ $N_2$ and 25 g of $H_{2}$ $H_2$ are reacted together, which one would be the limiting reactant?