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How many moles of ammonia are produced when 6 mol of hydrogen gas react with an excess of nitrogen gas?
Wiki User
∙ 13y ago
3H2 + N2 <------> 2NH3 is the balanced equation for Hydrogen and Nitrogen making ammonia. 3 moles of H2 produces two moles of ammonia and thus to make 6 moles requires 9 moles of Hydrogen.
Wiki User
∙ 13y ago
Wiki User
∙ 10y agoN2 + 3H2 ---> 2NH3 so 3 moles of hydrogen produces 2 moles of ammonia. Therefore with 6 moles of hydrogen available, 4 moles of ammonia only are possible as the hydrogen is the yield limiting material.
Wiki User
∙ 13y agoN2(g) + 3H2(g) --> 2NH3(g)
1 mol : 3 mol --> 2 mol
6 mol H2 (2 mol NH3 / 3 mol H2)
6(2/3) = 4 mol NH3
Wiki User
∙ 10y ago3H2+N2 --->2NH3 So you can double of moles N2.You can get 9.12mol NH3.
Wiki User
∙ 6y ago33 moles or nitrogen and 9 moles of hydrogen are needed.
Wiki User
∙ 6y agoIf you have 6 moles of N2 you can make theoretically 12 moles of NH3.
Elizabeth Sumpter
the answer is actually 3.3 mol :))
Wiki User
∙ 10y agoyes
Anonymous
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How many moles of ammonia gas are produced by the reaction of 3 mole of nitrogen gas with excess hydrogen gas N2 H2 ---- NH3?
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
If an excess of nitrogen is reacted with 3.41 grams of hydrogen how many grams of ammonia can be produced?
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
What is the difference between ammonia and ammonium chloride?
Ammonia is NH3, or a nitrogen atom bonded to 3 hydrogen atoms. By replacing one, two, or all three of the hydrogens with ethanol ( CH3-CH2-OH ), with the CH3 end being the one that bonds to the nitrogen, you create an ethanolamine. If one hydrogen is replaced with ethanol, you get monoethanolamine (MEA); if two hydrogens are replaced, you get diethanolamine (DEA); if all three hydrogens are replaced, you get triethanolamine (TEA). MEA can be represented as NH2-(CH2-CH2-OH). DEA can be represented as NH-(CH2-CH2-OH)2. TEA can be represented as N-(CH2-CH2-OH)3.
How many liters of nitrogen dioxide is produced when 34L of oxygen reacts with an excess of nitrogen monoxide?
Answer: 68 L NO22NO + O2 ------> 2NO2>68 L + 34 L --> 68 L + some 'left over' NO
How many grams of hydrogen must react if the reaction needs to produce 63.5 grams of ammonia?
Assume all of the nitrogen reacts to form ammonia, and it is all in the gaseous state. 1 N2 + 3 H2 --> 2 NH3. This means there are two moles of ammonia for every mole of N2., or 2 x 22.4 = 44.8 Liters of ammonia at STP for every 28 g of N2 So 56.0 g N2 will yield 89.6 L of ammonia gas. Your teacher probably wants the stoichiometry in the "railroad tracks", or "conversion factor" format where you start with the given, write a series of ratios that are all equal to 1, and multiply them together so all of the units drop out and you have the correct answer. This method will also help you catch your mistakes and solve many more difficult problems with ease (with a little practice). In this case: (56.0 g N2) x (1 mol N2/28g N2) X (2 mol NH3/1 mol N2) x (22.4 L NH3 at STP / mol NH3) = 89.6 L of ammonia gas at STP.
How many grams of NH3 can be produced from 4.12mol of N2 and excess H?
The gram molecular mass of ammonia is 17.03. The formula shows that only one atom of nitrogen is required for each mole of ammonia; 4.12 mol of diatomic nitrogen contains 8.24 mol of nitrogen atoms, and with excess H, all of this nitrogen can be converted to ammonia. Therefore, 8.24 mol of ammonia can be produced, and multiplying this number by17.03 yields a total mass of 140.3 grams of ammonia, to the justified number of significant digits.
How many moles of ammonia gas are produced by the reaction of 3 mole of nitrogen gas with excess hydrogen gas N2 H2 ---- NH3?
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
If an excess of nitrogen is reacted with 3.41 grams of hydrogen how many grams of ammonia can be produced?
Balanced equation first. N2 + 3H2 >> 2NH3 (hydrogen is limiting and drives the reaction ) 3.41 grams H2 (1mol/2.016g )(2mol NH3/3mol H2 )(17.034g NH3/1mol NH3 ) = 19.2 grams of ammonia produced ( this is called the Born-Haber process )
Are hydrogen gas and ammonia gas explosive?
Hydrogen is explosiveAmmonia when mixed with oxygen, it burns with a pale yellowish-green flame.At high temperature and in the presence of a suitable catalyst, ammonia is decomposed into its constituent elements. Ignition occurs when chlorine is passed into ammonia, forming nitrogen and hydrogen chloride; if chlorine is present in excess, then the highly explosive nitrogen trichloride (NCl3) is formed.
N2(g)+3H2(g)=2NH3(g)If a sample of 4.50 moles of hydrogen gas was reacted with excess nitrogen how many moles of ammonia gas would be produced?
3.50 W 21.80 x 1/14th
What do high temperature's high pressure combustion's nitrogen fixation's and lightning produce a excess carbon b excess nitrogen c nitrates d ammonia?
excess nitrogen
When acids in water what positive ion is produced in excess?
When acids in water hydrogen positive ion is produced in excess. It is this hydrogen positive ion that gives acidity of a solution.
If .3 mole of N2 are reacted with .6 mole of H2 what mass of NH3 could form the formula N2 plus 3 H2 makes 2 NH3 The answer to the problem is 6.8 but don't know how to get that?
First you have to find the limiting reactant. You have .3 moles of nitrogen and .6 moles of hydrogen, but you don't know which one is going to run out first.In any of these stoichiometry problems, you need to write down the formula:N2 + 3H2 → 2NH3Take both nitrogen and hydrogen and figure out how much ammonia is made alone..6 moles Hydrogen ÷ 3 moles hydrogen × 2 moles ammonia = .4 moles ammonia made.3 moles Nitrogen ÷ 1 mole nitrogen × 2 mole ammonia = .6 moles ammonia madeNow you figured out that hydrogen is the limiting reactant and the nitrogen is the excess because less ammonia is made using hydrogen. This measurement is what you will be using for the rest of the problem.Take the limiting reactant and use stoichiometry to find how much ammonia can be made.You could start with .6 moles of hydrogen and do the same conversion as above, but add the step of converting to grams. Or, since you already found out that .4 moles ammonia is made, just convert it to grams. The molecular mass of ammonia is 17.0 grams..4 moles ammonia × 17.0 grams = 6.8 grams ammonia
Kjeldahl method for estimation of nitrogen in inorganic organic compound?
This is a very convenient method but its use is restricted.This method is suitable for estimating nitrogen in those organic compounds in which nitrogen is linked to carbon and hydrogen. The method is not used in the case of nitro, azo and azoxy compounds. The method is extensively used for estimated nitrogen in food, fertilizers and agricultural products.Principle:- the method is based on the fact that when the nitrogenous compound is heated with concentrated sulphuric acid in presence of copper sulphate, the nitrogen present in the compound is quantitatively converted to ammonium sulphate. The ammonium sulphate so formed is decomposed with excess of alkali and the ammonia evolved is estimated volumetrically. The percentage of nitrogen is then calculated from the amount of ammonia.
How many grams of NH 3 can be produced from 3.91 mol of N 2 and excess H 2?
110 g ammonia are produced.
What is the difference between ammonia and ammonium chloride?
Ammonia is NH3, or a nitrogen atom bonded to 3 hydrogen atoms. By replacing one, two, or all three of the hydrogens with ethanol ( CH3-CH2-OH ), with the CH3 end being the one that bonds to the nitrogen, you create an ethanolamine. If one hydrogen is replaced with ethanol, you get monoethanolamine (MEA); if two hydrogens are replaced, you get diethanolamine (DEA); if all three hydrogens are replaced, you get triethanolamine (TEA). MEA can be represented as NH2-(CH2-CH2-OH). DEA can be represented as NH-(CH2-CH2-OH)2. TEA can be represented as N-(CH2-CH2-OH)3.
How many grams of NH3 can be produced from 2.23 mol of N2 and excess H2.?
76 g ammonia are obtained.
