**Step 1:** From the problem, the given combustion reaction is:

N_{2} _{(g)} + 3 H _{2} _{(g)} → 2 NH _{3} _{(g)}

Now that we have a balanced equation, we can proceed with the problem.

**Step 2:** Notice that we are given the mass of both reactants: this means we need to determine the ** limiting reactant**, which is the reactant that forms the less amount of product.

This is because once the limiting reactant is all used up, the reaction can no longer proceed and make more products.

This means the limiting reactant determines the **maximum mass of the product formed**.

We need to perform a ** mole-to-mole comparison** between each reactant and NH

**From N _{2}:**

molar mass N_{2} = 28.02 g/mol

molar mass NH_{3} = 17.034 g/mol

$\mathbf{mass}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}}{\mathbf{28}\mathbf{.}\mathbf{02}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{N}}_{\mathbf{2}}}}\mathbf{\times}\frac{\mathbf{17}\mathbf{.}\mathbf{034}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}$

**mass NH _{3} = 1215.8 g NH_{3}**

**From H _{2}:**

$\mathbf{mass}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\times}{\mathbf{10}}^{\mathbf{2}}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}{\mathbf{2}\mathbf{.}\mathbf{016}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}\mathbf{\times}\frac{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}{\mathbf{3}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}}\mathbf{\times}\frac{\mathbf{17}\mathbf{.}\mathbf{034}\mathbf{}\mathbf{g}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{NH}}_{\mathbf{3}}}}$

**mass NH _{3} = 2816.5 g NH_{3}**

Since N_{2} forms fewer NH_{3} → **H _{2} is the limiting reactant**.

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation:

N_{2} _{(g)} + 3 H _{2} _{(g)} → 2 NH _{3} _{(g)}

Given a mixture of 1.00 x 10 ^{3} g N_{2} and 5.00 x 10 ^{2} g H_{2}, what mass of which starting material would remain unreacted?

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Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. You can view video lessons to learn Limiting Reagent. Or if you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.

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Based on our data, we think this problem is relevant for Professor Sharma's class at Kean University.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.