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The balanced reaction for the Haber process that produces ammonia to feed the world will be used in the next four reaction stoichiometry problems. 3H_(2)+N_(2)longleftrightarrow 2NH_(3) How many grams of ammonia are produced when 28 grams of nitrogen gas are reacted with excess hydrogen gas? 17 grams 34 grams 2 grams 56 grams

Question

The balanced reaction for the Haber process that produces ammonia to feed the
world will be used in the next four reaction stoichiometry problems.
3H_(2)+N_(2)longleftrightarrow 2NH_(3)
How many grams of ammonia are produced when 28 grams of nitrogen gas are
reacted with excess hydrogen gas?
17 grams
34 grams
2 grams
56 grams

The balanced reaction for the Haber process that produces ammonia to feed the world will be used in the next four reaction stoichiometry problems. 3H_(2)+N_(2)longleftrightarrow 2NH_(3) How many grams of ammonia are produced when 28 grams of nitrogen gas are reacted with excess hydrogen gas? 17 grams 34 grams 2 grams 56 grams

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SiennaMaster · Tutor for 5 years

Answer

B. 34 grams

Explain

To solve this problem, we will follow these steps:1. Write down the balanced chemical equation:\[3 \mathrm{H}_{2} + \mathrm{N}_{2} \longleftrightarrow 2 \mathrm{NH}_{3}\]2. Calculate the molar mass of nitrogen gas (N2) and ammonia (NH3). The molar mass of nitrogen (N) is approximately 14 g/mol, so for N2, it would be:\[14 \, \text{g/mol} \times 2 = 28 \, \text{g/mol}\]The molar mass of ammonia (NH3) is calculated as follows:\[14 \, \text{g/mol} (for N) + 1 \, \text{g/mol} \times 3 (for H) = 14 \, \text{g/mol} + 3 \, \text{g/mol} = 17 \, \text{g/mol}\]3. Determine the number of moles of nitrogen gas used:\[\text{Moles of N2} = \frac{\text{mass of N2}}{\text{molar mass of N2}} = \frac{28 \, \text{g}}{28 \, \text{g/mol}} = 1 \, \text{mol}\]4. Use the stoichiometry of the balanced equation to find the moles of ammonia produced. According to the equation, 1 mole of N2 produces 2 moles of NH3.5. Calculate the mass of ammonia produced:\[\text{Mass of NH3} = \text{moles of NH3} \times \text{molar mass of NH3}\]\[\text{Mass of NH3} = 2 \, \text{moles} \times 17 \, \text{g/mol}\]\[\text{Mass of NH3} = 34 \, \text{g}\]
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