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Assuming you are talking about production from the Haber process, and also assuming Hydrogen is not the limiting quantity and also assuming 100% conversion, then the answer is as follows: N2 + 3H2 ----> 2NH3 Then 1 mole of N2 produces 2 moles of ammonia. So 3.94 moles of N2 produces 7.92 moles of ammonia. Taking the molecular weight of ammonia to be 17 then 17 x 7.92 = 134.64 g of ammonia produced.
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How many grams of NH3 will have the same number of molecules at 15.0 g of C6H6?
You first need to find the mass weight of NH3. wt. of N + (wt. of H)= 14.0067 + 3(1.0067)= 17.03052 Now that you have the mass weight, you divide 15 into 17.03052... 15.0/17.03052 = 0.8807 moles in 15. g of NH3
How many grams of ccl4 are needed to have 5.000 mol?
769.0 grams
How many grams of fructose are in 1.20 moles of fructose?
moles = weight in grams / molecular weight = 72 / 180 = 0.4 moles
What is the mass in grams of 1.00 mol Al?
7 X 102 g Al this is the awnser using significant figures
How do you Convert 86.2 grams c small 2 hsmal 4 to moles?
Carbon (C) has a atomic weight of 12 gr/mol Hydrogen (H) has an atomic weight of 1 gr/mol Therefore C2H4 has a molecular weight of 28 gr/mol (2x12 + 4x1) By dividing 86.2 grams by 28 gr/mol, you get 3.59 moles.
How many grams of NH3 can be produced from 2.08 grams of N2?
Ok, so I'm assuming that the chemical formula is written as - 3H2 + N2 ----> 2NH3 2.80 = moles of N2 17.03052 g/mol = Molar mass of NH3 (2.80 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 95.4 g NH3
N2 plus 3H2 -- 2NH3 If you produce 55.5 grams of ammonia how many grams of nitrogen will you need?
You need the balanced chemical equation for N2 and H2 combining to form ammonia, NH3.N2 (g) + 3 H2 (g) -----> 2 NH3 (g)Moles NH3 = ( 55.5 g NH3 ) / ( 17.03 g/mol NH3 ) = 3.259 moles of NH3n N2 required = ( 3.259 mol NH3 ) ( 1 N2 mol / 2 NH3 mol ) = 1.629 moles N2m N2 required = ( 1.629 mol N2 ) ( 28.103 g N2 / mol N2 ) = 45.67 g N2 needed
How many grams of NH3 can be produced from 4.33 mol of N2 and excess H2. Express your answer numerically in grams.?
The mass of ammonia is 147,5 g.
How many grams of NH3 can be produced from 2.23 mol of N2 and excess H2.?
76 g ammonia are obtained.
How many grams of NH3 can be produced from 2.63 mol of N2 and excess H2?
89,6 g ammonia are obtained.
What is the molarity of a solution that contains 17g of NH3 in 0.50l of solution?
Molarity = moles of solute/Liters of solution need to find moles NH3 16.7 grams NH3 (1 mole NH3/17.034 grams) = 0.9804 moles NH3 --------------------------------now Molarity = 0.9804 moles NH3/1.50 Liters = 0.654 M -------------
How many moles are in 1.2 X 10.3 grams of ammonia NH2?
Ammonia is NH3 (not NH2) so its molar mass is 17 g/mol. In 1.2*10^3 g there are1.2*10^3(g) / 17 (g/mol) = (70.6 =) 71 mol NH3.
How many grams of H2 are needed to produce 12.58 grams of NH3?
12.01 g NH3 = 0.667 mol NH3 = 1 mol H2 = 2.0 g H2
How many grams of oxygen are needed to react completely with 200.0 g of ammonia, NH3?
The balanced chemical reaction is: 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows: 200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed
How many grams are 4.5 moles of NH3?
Here is the solution,for 1mole NH3 or ammonia,N: 14.0067 g/mol x 1H: 1.00794 g/mol x 3So approximately it is--> 17.03052 g/mol)Now multiply this into 4.5 for 4.5 moles of ammonia, --> 4.5x17.03052 =76.64g
How many grams of NH3 can be produced from 4.10 moles of N and excess H?
There is no compound NH. However, there is ammonia, NH3. The reactants are nitrogen gas, N2, and hydrogen gas, H2.Multiply moles N2 by the mole ratio from the balanced equation between NH3 and N2, so that NH3 is in the numerator. Then multiply by the molar mass of NH3, 17.031 g/mol.Balanced equation: N2 + 3H2 --> 2NH34.10 mol N2 x (2 mol NH3)/(1 mol N2) x (17.031 g NH3)/(1 mol NH3) = 140. g NH3 = 1.40 x 10^2 g NH3 rounded to three significant figures.
How many mole are represented by 5.67 grams of CO 2 and 25.45 grams of NH3?
5.67g of CO2 divided by 44g/mol (CO2's molar mass) gives you 0.129 moles. 25.45g of NH3 divided by 17g/mol (NH3's molar mass) gives you 1.497 moles. To get the molar mass, visit the periodic table and add up the numbers of each constituent element.
