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Example · Example 9

Q.For the reaction N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g), calculate the mass of ammonia (NH3\text{NH}_3) that can be produced from 28 g28\ \text{g} of nitrogen gas, assuming hydrogen is present in excess.

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Balanced equation: N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g). Since H2\text{H}_2 is in excess, N2\text{N}_2 is the limiting reagent and controls how much NH3\text{NH}_3 forms.

Moles of N2\text{N}_2 taken =2828.02=0.999≈1.00 mol= \dfrac{28}{28.02} = 0.999 \approx 1.00\ \text{mol}

From the equation, 1 mol N21\ \text{mol}\ \text{N}_2 produces 2 mol NH32\ \text{mol}\ \text{NH}_3, so: …

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