Ah, what a lovely question! To find the number of molecules in 1 gram of NH3, we first need to know its molar mass, which is approximately 17 grams per mole. Then, we use Avogadro's number, which tells us there are about 6.022 x 10^23 molecules in one mole of a substance. By dividing 1 gram by the molar mass of NH3 and then multiplying by Avogadro's number, we find there are roughly 3.55 x 10^22 molecules in 1 gram of NH3.
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)
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
To find the mass of 1.75 x 10^24 molecules of NH3, you would first calculate the molar mass of NH3 (17.03 g/mol). Then, you would convert the number of molecules to moles by dividing by Avogadro's number (6.022 x 10^23 molecules/mol). Finally, multiply the number of moles by the molar mass to find the mass.
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
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
The answer is 1,57.10e27 molecules.
The mass is 9,6.10e-22 g for 34 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)
1 mole N2O3 = 76.0116g N2O3 (atomic weights x subscripts) 1 mole N2O3 = 6.022 x 1023 molecules N2O3 75g N2O3 x (1mol N2O3/76.0116g N2O3) = 0.99mol N2O3 0.99mol N2O3 x (6.022 x 1023 molecules N2O3/1mol N2O3) = 6.0 x 1023 molecules N2O3
This value is 21,85 g.
To find the mass of 1.75 x 10^24 molecules of NH3, you would first calculate the molar mass of NH3 (17.03 g/mol). Then, you would convert the number of molecules to moles by dividing by Avogadro's number (6.022 x 10^23 molecules/mol). Finally, multiply the number of moles by the molar mass to find the mass.
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 molar mass of ammonia is 17,031 g.17,031 g NH3-------------------1 mol500 g NH3-------------------------xx= 500/17,031 = 29,36 moles
1 g of ammonia (NH3) is equal to 0,059 mol.
Assuming that this ammonia gas is at STP, you can use Avogadro's number to gind the number of moles of gas:(387 x 1021 molecules) x (1 mol / 6.02x1023particles) x (17.03 g / 1 mol) =110 g NH3
NH3 Molecules = ( 8.1 x 10^20 H atoms ) ( 1 NH3 molecule / 3 H atoms ) NH3 Molecules = 2.7 x 10^20 NH3 molecules NH3 moles = ( NH3 molecules ) / ( N Avogadro ) NH3 moles = ( 2.7 x 10^20 NH3 molecules ) / ( 6.022 x 10^23 molecules / mole ) NH3 moles = 4.48 x 10^-4 NH3 moles <--------------
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