First we are going to find number of H2 moles form it mass
H2 moles=(2.37*10^-4)/2 = 1.185*10^-4moles
Since 3 moles of H2 makes 2 moles of NH3
then by using this ration, we can find number of moles in NH3
NH3 moles= (2 * 1.185*10^-4)/3 = 0.79*10^-4 moles
Finally, we find number of NH3 molecules by multiplying the number of moles with (6.022*10^23).
#NH3 molecules = (0.79*10^-4) *(6.022*10^23)
=4.75 * 10^19 molecules of NH3
Good luck
Enas
Balanced equation.
N2 + 3H2 --> 2NH3
1.4 moles H2 (2 moles NH3/3 moles H2)
= 0.93 moles NH3 produced
=======================
3.19*10^24
0,044 moles of NH3 can be produced.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
1 mole
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
0,044 moles of NH3 can be produced.
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
Balanced equation. N2 + 3H2 --> 2NH3 1.4 moles H2 (2 moles NH3/3 moles H2) = 0.93 moles NH3 produced =======================
How many moles of NH3 are produced when 1.2 mol of nitrogen reacts with hydrogen?
1 mole
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
0,3 moles of nitrogen reacted.
The reaction would be H2 + 3N2 ==>2NH3moles H2 used = 5.69104 g x 1 mole/2.00 = 2.84552 moles H2moles NH3 produced (assuming N2 is NOT limiting) = 2 moles NH3/mole H2 x 2.84552 moles H2 = 5.69104 moles NH3 producedMolecules of NH3 produced = 5.69104 moles x 6.02x10^23 molecules/mole = 3.4x10^24 molecules
Using the molar mass of nh3, we find that we have 2.5 moles of nh3. Since 3 moles of h2o are produced per 2 moles of nh3, we see that we will produce 3.75 moles of h2o. This is equivalent to around 3.79 g.
The answer is 1,57.10e27 molecules.
8 mol