How many grams of NO2 can be formed from the reaction between 28 grams of nitrogen and 64 gram of oxygen?
N2+2O2⟶ 2NO2
one mole of N2, 2 moles of O2>>>2 moles NO2=92 grams.
To determine the number of grams of NO2 that can be formed from the given reactants, we need to use stoichiometry, which involves balancing the chemical equation and using the molar masses of the elements involved.
1. Begin by balancing the chemical equation:
N2 + 2O2 ⟶ 2NO2
2. Calculate the molar masses of the reactants and products:
Molar mass of N2 = 2 x atomic mass of nitrogen
= 2 x 14.01 g/mol
= 28.02 g/mol
Molar mass of O2 = 2 x atomic mass of oxygen
= 2 x 16.00 g/mol
= 32.00 g/mol
Molar mass of NO2 = atomic mass of nitrogen + 2 x atomic mass of oxygen
= 14.01 g/mol + 2 x 16.00 g/mol
= 46.01 g/mol
3. Determine the number of moles of each reactant:
Number of moles of nitrogen = mass of nitrogen / molar mass of nitrogen
= 28 g / 28.02 g/mol
= 1 mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 64 g / 32.00 g/mol
= 2 mol
4. Use the balanced equation to find the limiting reactant:
Since the stoichiometry for the reaction indicates that 1 mole of nitrogen reacts with 2 moles of oxygen to form 2 moles of NO2, we can see that nitrogen is the limiting reactant. This means that the number of moles of NO2 formed will be determined by the number of moles of nitrogen available.
5. Calculate the number of moles of NO2 formed:
Number of moles of NO2 = (Number of moles of nitrogen) x (Number of moles of NO2 / Number of moles of nitrogen)
= 1 mol x (2 mol NO2 / 1 mol N2)
= 2 mol
6. Finally, calculate the mass of NO2 formed:
Mass of NO2 = number of moles of NO2 x molar mass of NO2
= 2 mol x 46.01 g/mol
= 92.02 g
Therefore, from 28 grams of nitrogen and 64 grams of oxygen, you can form 92.02 grams of NO2.