Balanced equation is N2 + 3H2 ==> 2NH3
3.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 present
moles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formed
molecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)
The number of ammonia molecules is 59 720.10e23.
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
The number of ammonia molecules is 5,83.10e19.
323.2 grams of KO2 is equal to 4.55 moles of KO2. Per the equation, 3.4125 moles of O2 is produced which is equal to 2.055 E24 molecules of oxygen. This means that every mg of KO2 produces 6.36 E18 molecules of O2.
6 moles COULD be produced
The answer is 1,57.10e27 molecules.
The number of ammonia molecules is 59 720.10e23.
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
4.8/16 moles of oxygen atoms converts to 1.6/16 moles of ozone molecules.
The number of ammonia molecules is 5,83.10e19.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
323.2 grams of KO2 is equal to 4.55 moles of KO2. Per the equation, 3.4125 moles of O2 is produced which is equal to 2.055 E24 molecules of oxygen. This means that every mg of KO2 produces 6.36 E18 molecules of O2.
6 moles COULD be produced
1,4 moles of CO are produced.
4.651024 molecules of NO2 equals 7,721 moles.
23 moles of oxygen contain 138,509.10e23 molecules.
The reaction requires 2 moles of hydrogen gas and 1 mole of oxygen gas to produce 2 moles of water.