Could someone please check my answers/work for the following questions.
EQUATION: 3H2+N2 -->2NH3
1) How many grams of NH3 can be produced from 3.98 mol of N2?
3.98 mol N2(2 mol NH3/2 mol H2)=2.65 mol
Convert to grams: 2.65(17.0)=45.14g
2.)How many grams of H2 are needed to produce 10.41g of NH3?
10.41g/17.0=0.61 mol
0.61 mol NH3(3mol H2/1mol N2)= 1.83 mol
(1.83)(2.016)=3.75g
3.)How many molecules of NH3 are produced from 4.98x10^-4g of H2?
(4.98X10^-4g/2.016)=2.47X10^-4mol
2.47x10^-4mol H2x (2mol NH3/1mol N2)=4.94 X10^-4 mol NH3(6.02X10^23)= 2.97 X 10^20 molecules.
Please let me know if I've done these questions correctly. Thanks
All are incorrect for the same reason. Everything is ok except the conversion factors.
1) How many grams of NH3 can be produced from 3.98 mol of N2?
3.98 mol N2(2 mol NH3/2 mol H2)=2.65 mol
Convert to grams: 2.65(17.0)=45.14g
3.98 mol N2 x (2 mol NH3/1 mol N2) = 7.96 mol NH3.
Then 7.96 mol NH3 x (17.0 g NH3/mol)= ?? g NH3.
Note the mole N2 cancel in my work to leave mol NH3 as the unit to keep. In your work, mol NH3 is the unit you keep (correct) BUT mol N2 in the numerator does not cancel with the mol H2 in the denominator. The other problems you did also have the same error. Otherwise your work is VERY good. My guess, since all were done the same way, is that you had a problem with your thought process. Just remember the factor used must cancel the unit you have and convert to the unit you want to keep.
H2(g)+ N2(g)NH3(g)
H2(g)+ N2(g)NH3(g) Please balance equation. Thanks
To verify your answers, we can use stoichiometry and the given balanced equation. Let's check each question:
1) How many grams of NH3 can be produced from 3.98 mol of N2?
Your calculation is correct:
3.98 mol N2 * (2 mol NH3 / 2 mol N2) = 3.98 mol of NH3
To convert from moles to grams, we multiply by the molar mass of NH3:
3.98 mol NH3 * (17.0 g NH3 / 1 mol NH3) = 67.66 g NH3
So the correct answer is 67.66 grams of NH3.
2) How many grams of H2 are needed to produce 10.41g of NH3?
Your calculation is correct:
10.41 g NH3 / (17.0 g NH3 / 1 mol NH3) = 0.61 mol NH3
Using the stoichiometry of the balanced equation, we can convert from moles of NH3 to moles of H2:
0.61 mol NH3 * (3 mol H2 / 1 mol NH3) = 1.83 mol H2
To convert from moles to grams, we multiply by the molar mass of H2:
1.83 mol H2 * (2.016 g H2 / 1 mol H2) = 3.68 g H2
So the correct answer is 3.68 grams of H2.
3) How many molecules of NH3 are produced from 4.98x10^-4g of H2?
Your calculation is correct:
4.98x10^-4 g H2 / (2.016 g H2 / 1 mol H2) = 2.47x10^-4 mol H2
Using the stoichiometry of the balanced equation, we can convert from moles of H2 to moles of NH3:
2.47x10^-4 mol H2 * (2 mol NH3 / 1 mol N2) = 4.94x10^-4 mol NH3
To convert from moles to molecules, we multiply by Avogadro's number (6.02x10^23 molecules/mol):
4.94x10^-4 mol NH3 * (6.02x10^23 molecules / 1 mol NH3) = 2.97x10^20 molecules of NH3
So the correct answer is 2.97x10^20 molecules of NH3.
Therefore, your answers are correct for all three questions! Well done!