what mass of ammonia nh3 is necessary to react with 2.1 x 10^24 molecules of oxygen when ammonia reacts with oxygen to form water and nitrogen dioxide
I need to do this using dimensional analysis
i dont know how to get this answer
To determine the mass of ammonia (NH3) necessary to react with a given number of oxygen (O2) molecules, we need to balance the chemical equation and then use stoichiometry.
The balanced equation for the reaction between ammonia and oxygen to form water and nitrogen dioxide is:
4 NH3 + 5 O2 -> 4 H2O + 4 NO2
From the balanced equation, we can see that 4 moles of NH3 react with 5 moles of O2.
1 mole of any substance contains Avogadro's number of molecules, which is approximately 6.022 x 10^23 molecules.
So, in 1 mole of O2, there are 6.022 x 10^23 molecules.
Now, we can calculate the number of moles of O2 that reacted using the given number of molecules.
Number of moles of O2 = (Number of molecules of O2) / (6.022 x 10^23 molecules/mol)
Number of moles of O2 = (2.1 x 10^24 molecules) / (6.022 x 10^23 molecules/mol)
Number of moles of O2 ≈ 3.48 moles
According to the stoichiometry of the balanced equation, 4 moles of NH3 react with 5 moles of O2.
So, if 5 moles of O2 react, we need to find out how many moles of NH3 are required.
Number of moles of NH3 = (Number of moles of O2) x (4 moles of NH3 / 5 moles of O2)
Number of moles of NH3 ≈ (3.48 moles) x (4/5) ≈ 2.784 moles
Finally, to find the mass of NH3, we can use its molar mass. The molar mass of NH3 is approximately 17.03 g/mol.
Mass of NH3 = (Number of moles of NH3) x (Molar mass of NH3)
Mass of NH3 ≈ (2.784 moles) x (17.03 g/mol) ≈ 47.49 g
Therefore, approximately 47.49 grams of ammonia (NH3) is necessary to react with 2.1 x 10^24 molecules of oxygen (O2).