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What mass of NH3 is produced when 1.20 mol of N2 react completely in the following equation N2 plus 3H2 2NH3?
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
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
Stoichiometry grams of N2 are needed to produce 20 grams NH3 N2 plus H2 NH3?
16,45 g nitrogen are needed.
How do you balaance N2 H2 --------------- NH3?
Put a 2 to NH3 first.Then a 3 to Hydrogen.
If 5.0 moles of NH3 are produce how many moles of N2 must have been used?
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
How do you balance N2 plus H2yeildsNH3?
(N2) + 3(H2) = 2(NH3)
If you have 14 grams of N2 how much NH3 can you make?
First you have to balance the equation N2 + H2 ---> NH3 N2 +3H2 ---> 2NH3 Then you have to use the Molecular Weight and number of mols required for complete reaction of each one to go from 14g N2 + xg of H2 to get the final result.
What is N2H4------NH3 plus N2 in a balanced equation?
3N2H4 --> 4NH3 + N2 is the correctly balanced equation.
What mass of NH3 is produced when 1.20 mol of N2 react completely in the following equation N2 plus 3H2 2NH3?
N2 + 3H2 ==> 2NH3moles N2 = 1.20 molesmoles NH3 formed = 1.20 moles N2 x 2 moles NH3/1 moles N2 = 2.40 moles NH3mass NH3 = 2.40 moles x 17 g/mole = 40.8 g NH3
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
How many grams of NH3 can be produced from 4.46 mol of N2 and excess H2.?
The molecular weight of NH3 is 17.03-grams per mole and 14.01 for N2. The reaction is N2 + 3H2 = NH3. Therefore for every 1-mole of N2 as a reactant 1-mole of NH3 is produced. .2941-moles of NH3 is produced with a mass of 5.01-grams.
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 N2 must react to form 1.7 grams of ammonia NH3?
To find out how many grams of N2 are needed to produce 1.7 grams of NH3, you need to look at the balanced chemical equation for the reaction. For the reaction N2 + 3H2 -> 2NH3, the molar ratio of N2 to NH3 is 1:2. So you would need half the number of grams of N2 as NH3, which is 0.85 grams of N2.
Stoichiometry grams of N2 are needed to produce 20 grams NH3 N2 plus H2 NH3?
16,45 g nitrogen are needed.
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 moles of N2 reacted if 0.40 mole NH3 is produced?
2 x 0.60 = 1.2 the reaction is N2 + 3H2 -> 2NH3 (1 mole of nitrogen N2 give 2 moles of NH3)
How do you balaance N2 H2 --------------- NH3?
Put a 2 to NH3 first.Then a 3 to Hydrogen.
