a. How many atoms of hydrogen are in 25.0 g of ammonia (NH3)? Use the format 6.02E+23 for 6.02 x 10^23
b. How many atoms of nitrogen are in the same amount of ammonia
I know to find the number of atoms in a gram of a sample you must multiply by 6.02*10^23.
So i multiplied 25*6.02*10^23
ammonias Molecular mass = 17.03
so do i divide 25.0/17.03=1.468 mol NH3
what next
now you have the moles of NH3.
then, there are three atoms of H per molecule, so moles of H is
3*1.468
number of H atoms
3*1.468*avagNumber
To find the number of atoms of hydrogen in 25.0 g of ammonia (NH3), you need to follow these steps:
1. Start by calculating the number of moles of NH3 in 25.0 g. Divide the mass of NH3 by its molar mass (17.03 g/mol).
25.0 g NH3 / 17.03 g/mol = 1.468 mol NH3
2. Next, use the balanced chemical equation of ammonia to determine the stoichiometry between NH3 and H2.
The balanced equation for the formation of ammonia is:
N2 + 3H2 → 2NH3
From the equation, you can see that 3 moles of hydrogen (H2) react to form 2 moles of ammonia (NH3).
3. Multiply the number of moles of NH3 (1.468 mol) by the stoichiometric ratio of H2 to NH3 (3/2).
1.468 mol NH3 * (3/2) = 2.202 mol H2
4. Finally, convert the number of moles of hydrogen (H2) to the number of atoms by multiplying by Avogadro's number (6.02 x 10^23 atoms/mol).
2.202 mol H2 * 6.02E+23 atoms/mol = 1.326E+24 atoms of hydrogen in 25.0 g of ammonia.
So, the answer to part (a) is 1.326E+24 atoms of hydrogen.
To find the number of atoms of nitrogen in the same amount of ammonia, you need to follow a similar process:
1. Start with the number of moles of NH3 from the previous step (1.468 mol NH3).
2. Use the stoichiometric ratio from the balanced chemical equation (1 mol N2 reacts with 2 mol NH3).
3. Multiply the number of moles of NH3 (1.468 mol) by the stoichiometric ratio of nitrogen (N2) to ammonia (NH3) (1/2).
1.468 mol NH3 * (1/2) = 0.734 mol N2
4. Convert the number of moles of nitrogen (N2) to the number of atoms by multiplying by Avogadro's number.
0.734 mol N2 * 6.02E+23 atoms/mol = 4.413E+23 atoms of nitrogen in 25.0 g of ammonia.
So, the answer to part (b) is 4.413E+23 atoms of nitrogen.