What mass of ammonia (NH3), is necessary to react with 2.1 x 1024 molecules of O2? 4 NH3 + 7 O2 = 6 H2O + 4 NO2?
To determine the mass of ammonia (NH3) required to react with a given number of molecules of O2, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is: 4 NH3 + 7 O2 → 6 H2O + 4 NO2
From the equation, we can see that 4 moles of NH3 react with 7 moles of O2.
1 mole of any substance contains 6.022 x 10^23 particles (Avogadro's number), which in this case would be molecules.
So, to determine the number of moles of O2 in 2.1 x 10^24 molecules, we can use the following conversion:
Number of moles of O2 = Number of molecules of O2 / Avogadro's number
= 2.1 x 10^24 / 6.022 x 10^23
= 3.49 moles
According to the stoichiometry of the balanced equation, 4 moles of NH3 react with 7 moles of O2. Therefore, we can set up a proportion to determine the number of moles of NH3 needed.
(4 moles NH3 / 7 moles O2) = (x moles NH3 / 3.49 moles O2)
Cross-multiplying and solving for x, we get:
x = (4 moles NH3 / 7 moles O2) × 3.49 moles O2
x = 1.994 moles NH3
Finally, we can convert moles of NH3 to mass using the molar mass of NH3.
Molar mass of NH3 = 14.01 g/mol + 3 × 1.01 g/mol (atomic masses of N and H)
= 17.03 g/mol
Mass of NH3 = moles of NH3 × molar mass of NH3
= 1.994 moles × 17.03 g/mol
= 33.92 g
Therefore, approximately 33.92 grams of ammonia (NH3) is necessary to react with 2.1 x 10^24 molecules of O2.
To find the mass of ammonia (NH3) necessary to react with a given number of oxygen molecules (O2), you first need to determine the ratio of ammonia to oxygen in the balanced chemical equation.
The balanced chemical equation is: 4 NH3 + 7 O2 = 6 H2O + 4 NO2
From the equation, you can see that 4 moles of ammonia react with 7 moles of oxygen to produce 6 moles of water and 4 moles of nitrogen dioxide.
Now, you need to convert the given number of oxygen molecules (O2) into moles.
1 mole of any gas contains 6.022 x 10^23 molecules (Avogadro's number).
Given: Number of oxygen molecules = 2.1 x 10^24
Number of moles of oxygen (O2) = (2.1 x 10^24) / (6.022 x 10^23) = 3.49 moles (approximately)
According to the balanced chemical equation, the mole ratio between ammonia (NH3) and oxygen (O2) is 4:7.
To find the number of moles of ammonia required, you can set up the following proportion:
4 moles of NH3 / 7 moles of O2 = x moles of NH3 / 3.49 moles of O2
Cross-multiplying, you get:
(4 moles of NH3) × (3.49 moles of O2) = (7 moles of O2) × (x moles of NH3)
13.96 = 7x
x = 13.96 / 7
x ≈ 1.99 moles (approximately)
Finally, to find the mass of ammonia (NH3) needed, you can use its molar mass, which is approximately 17.03 g/mol.
Mass of ammonia (NH3) = (1.99 moles) × (17.03 g/mol) = 33.86 grams (approximately)
Therefore, approximately 33.86 grams of ammonia (NH3) is necessary to react with 2.1 x 10^24 molecules of oxygen (O2).
well, 2.1*10^24 molecules = (2.1*10^24)/(6.02*10^23) = 3.488 moles of O2
4/7 that many moles of NH3 are needed
multiply that by the molar mass of NH3.