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
3H2 + N2 ----> 2NH3, right?
2.01 = moles of N2
17.03052 g/mol = Molar mass of NH3
(2.01 mol N2) x (2 NH3) / (1 N2) = 5.6 mol NH3 x (17.03052 g) / (1 mol NH3) = 68.463 g NH3
68,46 g ammonia are obtained.
95.0 g NH_3
78.0
83.79
160
11.6g
The answer is 20,664 g ammonia.
89,6 g ammonia are obtained.
0,044 moles of NH3 can be produced.
394.794 grams
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
The answer is 20,664 g ammonia.
The mass of ammonia will be 95,03 g.
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 )
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.
The mass of ammonia is 147,5 g.
76 g ammonia are obtained.
89,6 g ammonia are obtained.
The mass of nitrous oxide is 262,8 g.
0,044 moles of NH3 can be produced.
394.794 grams
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?