Hydrogen gas,H2 , reacts with nitrogen gas,N2 , to form ammonia gas,NH3 , according to the equation
3H2+N2-2NH3
How many grams ofNH3 can be produced from 2.90mol ofN2 ?
How many grams ofH2 are needed to produce 13.75g of NH3 ?
How many molecules (not moles) ofNH3 are produced from 8.26×10−4g ofH2 ?
3H2 + N2 ==> 2NH3
Convert 2.90 moles N2 to moles NH3 using the coefficients in the balanced equation.
2.90 x (2 moles NH3/1 mole N2) = 2.90 x 2/1 = ??
Convert moles NH3 to grams by g = moles x molar mass.
b.1. Convert 13.75 g NH3 to moles. moles = grams/molar mass = 13.75/17 = 1.057 moles NH3.
2. Convert moles NH3 to moles H2.
1.057 moles NH3 x (3 moles H2/2 moles NH3) = 1.585 moles H2.
3. Convert moles H2 to grams. 1.585 x 2 = 3.171 g to four significant figures (the number in 13.75 in the problem. You should go back through and use the proper molar masses to four s.f. since I just used these numbers from memory.
c. Start with 8.26 x 10^-4 g H2 and calculate moles of NH3. Then remember that 1 mole NH3 will contain 6.02 x 10^23 molecules.
Post your work if you need additional assistance.
To find the answers to these questions, we need to use the concept of stoichiometry and the balanced equation provided.
1) How many grams of NH3 can be produced from 2.90 mol of N2?
We start by looking at the balanced equation:
3H2 + N2 → 2NH3
The stoichiometry of the equation tells us that for every 1 mole of N2, we produce 2 moles of NH3. Therefore, we can set up a proportion equation:
2 mol NH3 / 1 mol N2 = x mol NH3 / 2.90 mol N2
Simplifying the equation, we find that x = (2 mol NH3 / 1 mol N2) x (2.90 mol N2) = 5.80 mol NH3.
Now, to convert the moles of NH3 to grams, we use the molar mass of NH3:
Molar mass of NH3 = 1 * molar mass of N + 3 * molar mass of H = 17.03 g/mol
Mass of NH3 = moles of NH3 * molar mass of NH3 = 5.80 mol * 17.03 g/mol = 98.77 g
Therefore, 98.77 grams of NH3 can be produced from 2.90 moles of N2.
2) How many grams of H2 are needed to produce 13.75 g of NH3?
Again, we start with the balanced equation:
3H2 + N2 → 2NH3
The stoichiometry of the equation tells us that for every 3 moles of H2, we produce 2 moles of NH3. Therefore, we can set up a proportion equation:
2 mol NH3 / 3 mol H2 = 13.75 g NH3 / x g H2
Simplifying the equation, we find that x = (3 mol H2 / 2 mol NH3) x (13.75 g NH3) = 20.63 g H2.
Therefore, 20.63 grams of H2 are needed to produce 13.75 grams of NH3.
3) How many molecules (not moles) of NH3 are produced from 8.26 × 10^-4 g of H2?
First, we need to convert grams of H2 to moles using the molar mass of H2:
Molar mass of H2 = 2.02 g/mol
Moles of H2 = mass of H2 / molar mass of H2 = (8.26 × 10^-4 g) / (2.02 g/mol) = 4.09 × 10^-4 mol
Now, using the balanced equation:
3H2 + N2 → 2NH3
We can calculate the number of moles of NH3 produced:
2 mol NH3 / 3 mol H2 = x mol NH3 / 4.09 × 10^-4 mol H2
Simplifying the equation, we find that x = (2 mol NH3 / 3 mol H2) x (4.09 × 10^-4 mol H2) = 2.73 × 10^-4 mol NH3
To convert moles to molecules, we use Avogadro's number:
1 mol = 6.022 × 10^23 molecules
Number of molecules of NH3 = moles of NH3 * Avogadro's number = (2.73 × 10^-4 mol) * (6.022 × 10^23 molecules/mol) = 1.64 × 10^20 molecules
Therefore, 1.64 × 10^20 molecules of NH3 are produced from 8.26 × 10^-4 g of H2.