Top 41 How Many Grams Of Nh3 Can Be Produced The 106 Correct Answer

How many grams of NH3 can be produced from 2.44 mol of N2 and excess H2? | Socratic

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• Summary of article content: Articles about How many grams of NH3 can be produced from 2.44 mol of N2 and excess H2? | Socratic Warning! Long Answer. (a) 83.1 g; (b) 2.264 g; (c) 3.17×1019lmolecules. Explanation: (a) Mass of ammonia. You know that you will need a … …
• Most searched keywords: Whether you are looking for How many grams of NH3 can be produced from 2.44 mol of N2 and excess H2? | Socratic Warning! Long Answer. (a) 83.1 g; (b) 2.264 g; (c) 3.17×1019lmolecules. Explanation: (a) Mass of ammonia. You know that you will need a … Warning! Long Answer. (a) 83.1 g; (b) 2.264 g; (c) 3.17 × 10^19color(white)(l)”molecules” > (a) Mass of ammonia You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc. (i). Assemble all the information in one place. M_text(r):color(white)(mm)28.01color(white)(m)2.016color(white)(mll) 17.03 color(white)(mmmmm)”N”_2+ color(white)(l)”3H”_2 → “2NH”_3 n”/mol”: color(white)(m)2.44 (ii). Calculate the moles of “NH”_3 “Moles of NH”_3 = 2.44 color(red)(cancel(color(black)(“mol N”_2))) × (“2 mol NH”_3)/(1 color(red)(cancel(color(black)(“mol N”_2)))) = 4.88 color(white)(l) “mol NH”_3 (iii). Calculate the mass of “NH”_3 “Mass of NH”_3 = 4.88 color(red)(cancel(color(black)(“mol NH”_3)))× (“17.03 g NH”_3)/(1 color(red)(cancel(color(black)(“mol NH”_3)))) = “83.1 g NH”_3 (b) Mass of “H”_2 (i). Calculate the moles of “NH”_3 “Moles of NH”_3 = 12.75 color(red)(cancel(color(black)(“g NH”_3))) × “1 mol NH”_3/(17.03 color(red)(cancel(color(black)(“g NH”_3)))) = “0.748 68 mol NH”_3 (ii). Calculate the moles of “H”_2 “Moles of H”_2 = “0.748 68” color(red)(cancel(color(black)(“mol NH”_3))) × “3 mol H”_2/(2 color(red)(cancel(color(black)(“mol NH”_3)))) = “1.1230 mol H”_2 (iii). Calculate the mass of “H”_2 “Mass of H”_2 = 1.1230 color(red)(cancel(color(black)(“mol H”_2))) × “2.016 g H”_2/(1 color(red)(cancel(color(black)(“mol H”_2)))) = “2.264 g H”_2 (c). Molecules of “NH”_3 (i). Calculate the moles of “H”_2 “Moles of H”_2 = 1.59 × 10^”-4″ color(red)(cancel(color(black)(“g H”_2))) × “1 mol H”_2/(2.016 color(red)(cancel(color(black)(“g H”_2)))) = 7.887 × 10^”-5″color(white)(l) “mol H”_2 (ii). Calculate the moles of “NH”_3 ” Moles of NH”_3 = 7.887 × 10^”-5″ color(red)(cancel(color(black)(“mol H”_2))) × “2 mol NH”_3/(3 color(red)(cancel(color(black)(“mol H”_2)))) = 5.258 ×10^”-5″color(white)(l) “mol NH”_3 (iii). Calculate the molecules of “NH”_3 “Molecules of NH”_3 = 5.258 ×10^”-5″ color(red)(cancel(color(black)(“mol NH”_3))) × (6.022 × 10^23 color(white)(l)”molecules NH”_3)/( 1 color(red)(cancel(color(black)(“mol NH”_3)))) = 3.17 × 10^19color(white)(l)”molecules NH”_3#
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“How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2? N2

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• Summary of article content: Articles about “How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2? N2 grams NH3 = 2.0 x 17 = 34 g. moles H2 = 25/2 = 12.5 moles. moles NH3 produced = 12.5 x (2 moles NH3/3 moles H2) = 8.33 … …
• Most searched keywords: Whether you are looking for “How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2? N2 grams NH3 = 2.0 x 17 = 34 g. moles H2 = 25/2 = 12.5 moles. moles NH3 produced = 12.5 x (2 moles NH3/3 moles H2) = 8.33 …
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• Summary of article content: Articles about brainly.com Grams of NH3 = moles of NH3 * molar mass of NH3 = 8.16 mole * 17 g/mole = 139 g. 139 grams of nh3 can be produced from 4.08 mol of n2 and excess h2. …
• Most searched keywords: Whether you are looking for brainly.com Grams of NH3 = moles of NH3 * molar mass of NH3 = 8.16 mole * 17 g/mole = 139 g. 139 grams of nh3 can be produced from 4.08 mol of n2 and excess h2.
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• Summary of article content: Articles about brainly.in How many grams of NH3 can be prepared from 77.3 grams of N2 and 14.2 grams of H2?(Hint: N, +3H, → 2NH3) 80.4 g 79.7 g 47.0 g 120.0 g​ · Thus, … …
• Most searched keywords: Whether you are looking for brainly.in How many grams of NH3 can be prepared from 77.3 grams of N2 and 14.2 grams of H2?(Hint: N, +3H, → 2NH3) 80.4 g 79.7 g 47.0 g 120.0 g​ · Thus, …
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how many gram of nh3 will be produce when 50g n2 react with h2 – Chemistry – TopperLearning.com | t0qegjww

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How many grams of NH3 can be produced from 4.84 mol of N2? Access 14 best answers & solutions.

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• Summary of article content: Articles about How many grams of NH3 can be produced from 4.84 mol of N2? Access 14 best answers & solutions. Here are 14 best answers to ‘How many grams of NH3 can be produced from 4.84 mol … N2 + 2H2 –> 2NH3 As you can see that the mole ratio is N2 : NH3 1 : 2 … …
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Answer in Inorganic Chemistry for aleeyah #127447

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What is the molar mass of ammonia, #”NH”_3#?

To determine the molar mass of any compound, all you have to do is add up the molar masses of every atom that makes up the respective compound.

In this case, you know that ammonia, #”NH”_3# , is composed of

one nitrogen atom, #”N”#

nitrogen atom, three hydrogen atoms, #”H”#

This means that its molar mass will be the sum of the molar mass of one nitrogen atom and three times the molar mass of a hydrogen atom.

A quick look in the periodic table will show that you have

#”N: ” “14.0067 g/mol”#

#”H: ” “1.00794 g/mol”#

The molar mass of ammonia will thus be

How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

Part A: How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

Part B: How many grams of NH3 can be produced from 3.60 mol of N2 and excess H2?

Part C: How many grams of H2 are needed to produce 11.76g of NH3?

Part D: How many molecules (not moles) of NH3 are produced from 8.86 * 10^-4g of H2?

Concepts and reason

The question is based on the concept of limiting reagent.

Limiting reagent is a reagent in chemical reaction that gets totally consumed when chemical reaction is complete.

Fundamentals

There are two methods to determine the limiting reagent. One method is to find the mole ratio and then compare the mole ratio of reactants that get used in the reaction. Other method is to calculate the grams of products that are produced from the given quantities of reactants.

The mass of a substance can be calculated by using number of moles and molar mass of that substance as follows.

m = nM

Here, n is number of moles and M is molar mass of the substance.

Answer

Part A

2.10 moles of H2reacted with excess of N2. So H2 is a limiting reagent. The balanced chemical reaction can be written as follows:

The balanced chemical equation shows that 3 moles of H2 gives 2 moles ofNH3.

Part A

The moles of NH3 produced are 14 moles.

First write the balanced chemical equation. Then, solve the problem using unitary method.

Part B

2360 moles of N2reacted with excess of H2. So N2 is a limiting reagent. The balanced chemical reaction can be written as follows:

Now, mass of ammonia is calculated by using formula as follow.

Part B

The mass of NH3 produced is 122.6 g.

Part C

Write the balanced chemical reaction as follows:

Part C

The mass of H2 needed to produce 11.76 g of NH3 is 2.10 g.

First write the balanced chemical equation. Then, solve the problem using unitary method. After finding the moles of ammonia produced next step is to find the moles of hydrogen produced and with the help of this mass of hydrogen will be calculated.

Part D

Write the balanced chemical reaction as follows:

Part D

Molecules of NH3 produced are

First write the balanced chemical equation. Then, solve the problem using unitary method. After finding the moles of hydrogen produced next step is to find the molecules of ammonia produced.

N2(g) + 3H2-> 2NH3(g) How many grams of ammonia (NH3) can be produced from the reaction of 28.0 g of N2 and 25.0 g of H2?

N2(g) + 3H2-> 2NH3(g) This is the balanced equation

Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.

moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present

moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present

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

NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.

So you have finished reading the how many grams of nh3 can be produced topic article, if you find this article useful, please share it. Thank you very much. See more: how many grams of nh3 can be produced from 12.0 g of h2, how many grams of nh3 can be produced from 4.09 mol of n2 and excess h2, if you have 3.64 g of h2, how many grams of nh3 can be produced, how many grams of nh3 can be prepared from 77.3 grams of n2, how many grams of nh3 will be produced if 122 g n2 reacts completely with h2?, how many grams of ammonia nh3 are produced in the reaction with 50.0 g of n2 nitrogen, how many grams of nh3 can be produced from the reaction of 28 g of n2 and 25 g of h2, how many grams of nh3 can be produced from 3.48 mol of n2 and excess h2