# Ideal answer for — how many moles of NH3 can be produced from the reaction of 28g of N2?

Contents

One mole of N2 can produce two moles of NH3, therefore 28g of N2 can produce 1.004 moles of NH3.

## For those who want further information

According to the balanced chemical equation N2 + 3H2 → 2NH3, one mole of N2 reacts with three moles of H2 to produce two moles of NH3. Therefore, in order to find out how many moles of NH3 can be produced from the reaction of 28g of N2, we need to first convert the given mass of N2 into its corresponding number of moles.

To do this, we can use the molar mass of N2, which is approximately 28 g/mol. Dividing the given mass of N2 (28g) by its molar mass (28g/mol) gives us one mole of N2. Since one mole of N2 can produce two moles of NH3, multiplying the number of moles of N2 by the mole ratio of NH3 to N2 (2/1) gives us the number of moles of NH3 that can be produced.

Therefore, the calculation would be:

28g N2 ÷ 28 g/mol N2 = 1 mol N2

1 mol N2 × 2 mol NH3/1 mol N2 = 2 mol NH3

So, 28g of N2 can produce 2 moles of NH3.

As well-known scientist Marie Curie once said, “Nothing in life is to be feared, it is only to be understood. Now is the time to understand more, so that we may fear less.” Continuing to learn and understand the principles of chemistry can help us better comprehend the world around us.

Interesting facts about NH3:

• NH3, or ammonia, is a colorless gas that is commonly used in industrial processes such as refrigeration, fertilizer production, and the manufacturing of textiles and cleaning products.
• Ammonia is produced naturally in the environment through the decay of organic matter and the breakdown of nitrogen-containing compounds.
• In its pure form, ammonia is toxic and can be harmful if inhaled, ingested, or exposed to the skin or eyes.
• NH3 has a characteristic pungent odor that can be detected even in very low concentrations. It is often described as smelling like rotten eggs or urine.
• Ammonia is also used in household cleaning products such as window cleaners, floor waxes, and furniture polishes.
• NH3 is a common pollutant in the atmosphere, especially in areas with high levels of agriculture and livestock production. It can contribute to the formation of acid rain and other harmful environmental effects.
• When dissolved in water, ammonia forms ammonium ions (NH4+), which are important nutrients for plants. Fertilizers that contain ammonium ions are commonly used to enhance crop growth.
• The Haber-Bosch process is a method for synthesizing ammonia from nitrogen and hydrogen gases. This process was developed in the early 20th century and is still widely used today to produce ammonia for use in fertilizers and other applications.
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Table:

Property Value
Chemical formula NH3
Molar mass 17.03 g/mol
Melting point -77.73 °C
Boiling point -33.34 °C
Density 0.617 g/L (at STP)
Solubility in water Miscible
pKa 9.24
Odor Pungent, ammonia-like

Answer in the video

This video provides a detailed guide to solve a stoichiometry problem using the GWR (Given, Want, and Ratios) method. The problem involves producing ammonia from hydrogen gas and excess nitrogen, and the presenter illustrates the step-by-step process of identifying the given and want, using mole to mole ratios, finding ratios involving grams of hydrogen, and solving for the number of moles of ammonia produced. The narrator also suggests alternative methods to avoid calculation errors, such as hitting the division sign for each denominator or using parentheses to multiply.

## See more answer options

2 moles NH3 = 2 moles

How many moles of NH3 can be produced from the reaction of 28 g of N2 ? 28 g N2 X 1 mole N2 X 2 moles NH3 = 2 moles NH3

How many moles of NH3 can be produced from the reaction of 28 g of N2 ? 28 g N2 X 1 mole N2 X 2 moles NH3 = 2 moles NH3

Answer:34 g of NH₃ will be produced from the reaction of 28 g of N₂ with 25 g of H₂.Explanation:1) Balanced chemical equationN₂ (g) + 3H₂ (g) → 2NH₃(g)2) Stoichiometric (theoretical ) mole ratios1 mol N₂ (g) : 3mol H₂ (g) : 2 mol NH₃(g)3) Number of moles of each reactantnumber of moles = mass in grams / atomic massnumber of moles of N₂ = 28 g / 28 g/mol = 1 molnumber of moles of H₂: 25 g / 2 g/mol = 12.5 mol4) Limiting reactantSince the stoichiometry states that 1 mol of N₂ reacts with 3 moles of H₂, the given mass of N₂ will react completely with the given amount of H₂, and the calculations must be done with the 28 g (1 mol) of N₂ as the limiting reactant.5) YieldSet the proportion with the mole ratios:1 mol H₂ / 2 mol NH₃ = 1 mol H₂ / x ⇒ x = 2 mol NH₃6) Convert to gramsmass in grams = number of moles × molar mass = 2 mol × 17 g/mol = 34 g.Answer: the reaction of 28 g of N₂ with 25 g of H₂ will produce 34 g of NH₃

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### Which one would be the limiting reactant if 28 g of N2 and 25 g of H2 are reacted together?

As a response to this: 3. If 28 g of N2 and 25 g of H2 are reacted together, which one would be the limiting reactant? N2 would be the limiting reactant because it only makes 2 moles of NH3 before it is used up.

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### What is the limiting reactant of N2 g )+ 3h2 G → 2nh3 G?

Response: N2 (g)+ 3H2 (g) → 2NH3 (g) If equals moles of each reactant are used, hydrogen will be the limiting reagent as 3 moles are hydrogen are used up for every one mole of nitrogen to form ammonia.

### Which is the limiting reactant if equal grams of H2 and Cl2 react?

The reply will be: Chlorine
Chlorine, therefore, is the limiting reactant and hydrogen is the excess reactant (Figure 4.13). Figure 4.13 When H2 and Cl2 are combined in nonstoichiometric amounts, one of these reactants will limit the amount of HCl that can be produced.

### How many grams of excess reactant remain?

The answer is: To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.

### How many moles of NH3 would be produced if H2 and N2 react?

Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H2 + N2 2 NH3. If 4.0 moles of H2 with 2.0 mol of N2 are reacted how many moles of NH3 would be produced? Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H2 + N2 2 NH3.

### How many grams of NH3 can be produced?

moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

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### What is the limiting reactant of NH3?

Hydrogen is the limiting reactant. In other words, Hydrogen will run out before Nitrogen because it takes 3 mole of H2 and just 1 mole of N2 to make 2 mole NH3. How many grams of NH3 are produced by 4 moles of N2 and 10 moles of H2? Originally Answered: How many grams of NH3 is produced by 4 mole of N2 and 10 mole of H2? N2 is present in excess.

### Why is NH3 limiting based on mole ratio?

In reply to that: Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2. moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced

### How many moles of NH3 would be produced if H2 and N2 react?

Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H2 + N2 2 NH3. If 4.0 moles of H2 with 2.0 mol of N2 are reacted how many moles of NH3 would be produced? Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H2 + N2 2 NH3.

### What is the limiting reactant of NH3?

Answer to this: Hydrogen is the limiting reactant. In other words, Hydrogen will run out before Nitrogen because it takes 3 mole of H2 and just 1 mole of N2 to make 2 mole NH3. How many grams of NH3 are produced by 4 moles of N2 and 10 moles of H2? Originally Answered: How many grams of NH3 is produced by 4 mole of N2 and 10 mole of H2? N2 is present in excess.

### How many grams of NH3 can be produced?

As a response to this: moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

### How many Mol s of NH3 in 2 0.9995?

Response to this: We have 1 mol N2, and we need three times as many mol s of H2 as we have N2. After doing the actual calculation you should realize that we have about 4 times as much H2 as we need. Therefore the limiting reagent is clearly N2. Thus, we should yield 2 ×0.9995 = 1.9990 mol s of NH3 (refer back to the reaction).

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