The number of ammonia molecules is 5,83.10e19.
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 59 720.10e23.
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.
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules 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)
2 KClO3 ----> 2KCl + 3O2 So 2 moles of Potassium Chlorate produces 3 moles of oxygen molecules or 6 moles of oxygen atoms. 3 moles of Potassium chlorate would thus produce 4.5 moles of oxygen molecules or 9 moles of oxygen atoms.
The answer is 1,57.10e27 molecules.
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 59 720.10e23.
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.
4.8/16 moles of oxygen atoms converts to 1.6/16 moles of ozone molecules.
Balanced equation is N2 + 3H2 ==> 2NH33.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 presentmoles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formedmolecules 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)
2 KClO3 ----> 2KCl + 3O2 So 2 moles of Potassium Chlorate produces 3 moles of oxygen molecules or 6 moles of oxygen atoms. 3 moles of Potassium chlorate would thus produce 4.5 moles of oxygen molecules or 9 moles of oxygen atoms.
Given the balanced equation2C3H8O + 9O2 --> 6CO2 + 8H2OTo find the number of moles CO2 that will be produced from 0.33 mol C3H8O, we must convert from moles to moles (mol --> mol conversion).0.33 mol C3H8O * 6 molecules CO2 = 0.99 mol CO2---------- 2 molecules C3H8O
The reaction requires 2 moles of hydrogen gas and 1 mole of oxygen gas to produce 2 moles of water.
6 moles COULD be produced
1,4 moles of CO are produced.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?