0
I actually have this problem in my chemistry book too. This is what is given in the "Solutions to Red Exercises" book that went along with the text:
Analyze. Given: mol NH3, Find: N atoms
Plan. mol NH3---> mol N atoms---> N atoms
Solve. 0.410 mol NH3 * 1 mol N atoms/ 1 mol NH3 * 6.022*10^23 atoms/1mol = 2.47*10^23 N atoms
Check. (0.4)(6*10^23)=2.4*10^23.
I hope this helps.
What else can I help you with?
How many atoms are there in 170 grams of ammonia?
To calculate the number of atoms in 170 grams of ammonia (NH3), you would first determine the number of moles using the molar mass of ammonia (17.03 g/mol). Then, using Avogadro's number (6.022 x 10^23 atoms/mol), you find that there are approximately 1.79 x 10^24 atoms in 170 grams of ammonia.
How many grams of NH3 can be produced from 3.38 mol of N2 and excess H2.?
The balanced chemical equation for the reaction is: N2 + 3H2 -> 2NH3 From the equation, it can be seen that 1 mol of N2 produces 2 mol of NH3. Therefore, 3.38 mol of N2 will produce 2 x 3.38 = 6.76 mol of NH3. To convert this to grams, you need to multiply the molar mass of NH3 (17.03 g/mol) by the number of moles of NH3 produced. Thus, 6.76 mol of NH3 will produce 6.76 x 17.03 = 115.18 g of NH3.
What is the bond energy of nh3?
The bond energy of NH3, which is the energy required to break one mole of NH3 molecules into its individual atoms, is approximately 391 kJ/mol.
How many moles of NH3 are present in 107.1g?
To find the number of moles of NH3 in 107.1g, divide the given mass by the molar mass of NH3. The molar mass of NH3 is 17.03 g/mol. ( \frac{107.1 , \text{g}}{17.03 , \text{g/mol}} ≈ 6.29 , \text{mol} ) of NH3 are present.
How many atoms are in 63.9g of ammonia?
Ammonia = NH3 and has a molar mass of 17.031 g/molmoles NH3 = 63.9 g x 1 mol/17.031 g = 3.752 molesmolecules NH3 = 3.752 moles x 6.02x10^23 molecules/mole = 2.259x10^24 moleculesEach molecule of NH3 has 4 atoms (1 N + 3 H), thus....number of atoms = 4 atoms/molecule x 2.259x10^24 molecules = 9.04x10^24 atoms (to 3 sig figs)
How many atoms are there in 0.850 mol of ammonia?
There are 6.02 x 1023 NH3 molecules in 1 mole of NH3.Therefore 0.850mol of ammonia would have 0.85 x 6.02 x 1023 molecules of NH3.In each NH3 molecule there are 4 atoms (one N and three H atoms).Therefore the number of atoms in 0.850mol of NH3 is4 x 0.85 x 6.02 x 1023 = 2.05 x 1024
What is the molecular mass of NH3 when The atomic mass of N is 1401 gmole and the atomic mass of H is 1008 gmole?
Molecular mass = sum of all atoms masses = 1(molN/mol NH3)*14.01(g/mol N) + 3(molH/mol NH3)*1.008(g/mol H) = 17.03 g/mol NH3
How many atoms are there in 170 grams of ammonia?
To calculate the number of atoms in 170 grams of ammonia (NH3), you would first determine the number of moles using the molar mass of ammonia (17.03 g/mol). Then, using Avogadro's number (6.022 x 10^23 atoms/mol), you find that there are approximately 1.79 x 10^24 atoms in 170 grams of ammonia.
How many grams of NH3 can be produced from 3.38 mol of N2 and excess H2.?
The balanced chemical equation for the reaction is: N2 + 3H2 -> 2NH3 From the equation, it can be seen that 1 mol of N2 produces 2 mol of NH3. Therefore, 3.38 mol of N2 will produce 2 x 3.38 = 6.76 mol of NH3. To convert this to grams, you need to multiply the molar mass of NH3 (17.03 g/mol) by the number of moles of NH3 produced. Thus, 6.76 mol of NH3 will produce 6.76 x 17.03 = 115.18 g of NH3.
What is the bond energy of nh3?
The bond energy of NH3, which is the energy required to break one mole of NH3 molecules into its individual atoms, is approximately 391 kJ/mol.
How many moles of NH3 are present in 107.1g?
To find the number of moles of NH3 in 107.1g, divide the given mass by the molar mass of NH3. The molar mass of NH3 is 17.03 g/mol. ( \frac{107.1 , \text{g}}{17.03 , \text{g/mol}} ≈ 6.29 , \text{mol} ) of NH3 are present.
How many atoms are present in 80.0 mol of zirconium?
To determine the number of atoms in 80.0 mol of zirconium, you can use Avogadro's number, which is 6.022 x 10^23 atoms/mol. Multiply 80.0 mol by Avogadro's number to find the total number of atoms in 80.0 mol of zirconium.
How many grams of rm NH 3 can be produced from 2.90 mol of rm N 2?
To calculate the amount of NH3 that can be produced, we need to use the balanced chemical equation for the reaction between N2 and H2 to form NH3. The balanced equation is: N2 + 3H2 -> 2NH3. From the equation, we can see that 1 mole of N2 produces 2 moles of NH3. Therefore, if 2.90 mol of N2 is used, it will produce 2.90 mol * (2 mol NH3/1 mol N2) = 5.80 mol of NH3. To convert this to grams, we need to multiply the number of moles of NH3 by its molar mass, which is 17 g/mol. Therefore, the amount of NH3 produced from 2.90 mol of N2 is 5.80 mol * 17 g/mol = 98.6 g of NH3.
How many atoms are in 63.9g of ammonia?
Ammonia = NH3 and has a molar mass of 17.031 g/molmoles NH3 = 63.9 g x 1 mol/17.031 g = 3.752 molesmolecules NH3 = 3.752 moles x 6.02x10^23 molecules/mole = 2.259x10^24 moleculesEach molecule of NH3 has 4 atoms (1 N + 3 H), thus....number of atoms = 4 atoms/molecule x 2.259x10^24 molecules = 9.04x10^24 atoms (to 3 sig figs)
What is the answer. Determine the number of atoms in 3.29 mol Zn?
To determine the number of atoms in 3.29 mol of Zn, you would use Avogadro's number, which is 6.022 x 10^23 atoms/mol. Multiply the number of moles (3.29 mol) by Avogadro's number to find the number of atoms. The calculation would be: 3.29 mol Zn x 6.022 x 10^23 atoms/mol = 1.98 x 10^24 atoms of Zn.
How many grams of NH3 can be produced from 4.46 mol of N2 and excess H2.?
The molecular weight of NH3 is 17.03-grams per mole and 14.01 for N2. The reaction is N2 + 3H2 = NH3. Therefore for every 1-mole of N2 as a reactant 1-mole of NH3 is produced. .2941-moles of NH3 is produced with a mass of 5.01-grams.
How many grams of NH3 can be produced from 4.12mol of N2 and excess H?
The gram molecular mass of ammonia is 17.03. The formula shows that only one atom of nitrogen is required for each mole of ammonia; 4.12 mol of diatomic nitrogen contains 8.24 mol of nitrogen atoms, and with excess H, all of this nitrogen can be converted to ammonia. Therefore, 8.24 mol of ammonia can be produced, and multiplying this number by17.03 yields a total mass of 140.3 grams of ammonia, to the justified number of significant digits.
