What mass of water, #H_2O#, can be produced from 150 grams of ammonia, #NH_3#?

Question

Henry Antrim · Accepted Answer

238 g

#NH_3+3/4O_2 =1/2N_2+3/2H_2O# As guided by the above equation, we see 1 mole #"NH"_3# produces 1.5 moles of water. Hence #"150 g NH"_3 =150/17 "mol NH"_3# produces #150/17*1.5# moles #"H"_2"O"# which will have a mass #150/17*1.5 × "molar mass of H"_2"O"# #=150/17*1.5*18=238#

Noah Aldrich · Answer

undefined To calculate the mass of water produced from 150 grams of ammonia (NH3), we first need to write the balanced chemical equation for the reaction. The reaction is:

2NH3 (g) → H2O (l) + N2 (g)

From the balanced equation, we can see that 2 moles of ammonia (NH3) react to form 1 mole of water (H2O).

1 mole of NH3 has a molar mass of 17.031 g/mol.
1 mole of H2O has a molar mass of 18.015 g/mol.

Now, we can calculate the moles of NH3 in 150 grams:

$ \text{Moles of NH}_3 = \frac{150 \, \text{g}}{17.031 \, \text{g/mol}} $

Then, we use the mole ratio from the balanced equation to find the moles of H2O produced:

$ \text{Moles of H}_2\text{O} = \frac{\text{Moles of NH}_3}{2} $

Finally, we convert the moles of H2O to grams:

$ \text{Mass of H}_2\text{O} = \text{Moles of H}_2\text{O} \times 18.015 \, \text{g/mol} $

Now we calculate:

$ \text{Moles of NH}_3 = \frac{150 \, \text{g}}{17.031 \, \text{g/mol}} = 8.81 \, \text{mol} $

$ \text{Moles of H}_2\text{O} = \frac{8.81 \, \text{mol}}{2} = 4.41 \, \text{mol} $

$ \text{Mass of H}_2\text{O} = 4.41 \, \text{mol} \times 18.015 \, \text{g/mol} = 79.5 \, \text{g} $

So, 79.5 grams of water (H2O) can be produced from 150 grams of ammonia (NH3).