Question

How many grams of NH3 can be produced from 2.44 mol of N2 and excess H2?

How many grams of NH3 can be produced from 2.44 mol of N2 and excess H2.?
How many grams of H2 are needed to produce 12.75 g of NH3?
How many molecules (not moles) of NH3 are produced from 1.59×10−4 g of H2?

Answer 1

Warning! Long Answer. (a) 83.1 g; (b) 2.264 g; (c) `3.17 × 10^19color(white)(l)"molecules"`

Explanation:

(a) Mass of ammonia

You know that you will need a balanced chemical equation with masses, moles, and molar masses, etc.

(i). Assemble all the information in one place.

`M_text(r):color(white)(mm)28.01color(white)(m)2.016color(white)(mll) 17.03`
`color(white)(mmmmm)"N"_2+ color(white)(l)"3H"_2 → "2NH"_3`
`n"/mol": color(white)(m)2.44`

(ii). Calculate the moles of `"NH"_3`

`"Moles of NH"_3 = 2.44 color(red)(cancel(color(black)("mol N"_2))) × ("2 mol NH"_3)/(1 color(red)(cancel(color(black)("mol N"_2)))) = 4.88 color(white)(l) "mol NH"_3`

(iii). Calculate the mass of `"NH"_3`

`"Mass of NH"_3 = 4.88 color(red)(cancel(color(black)("mol NH"_3)))× ("17.03 g NH"_3)/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "83.1 g NH"_3`

(b) Mass of `"H"_2`

(i). Calculate the moles of `"NH"_3`

`"Moles of NH"_3 = 12.75 color(red)(cancel(color(black)("g NH"_3))) × "1 mol NH"_3/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "0.748 68 mol NH"_3`

(ii). Calculate the moles of `"H"_2`

`"Moles of H"_2 = "0.748 68" color(red)(cancel(color(black)("mol NH"_3))) × "3 mol H"_2/(2 color(red)(cancel(color(black)("mol NH"_3)))) = "1.1230 mol H"_2`

(iii). Calculate the mass of `"H"_2`

`"Mass of H"_2 = 1.1230 color(red)(cancel(color(black)("mol H"_2))) × "2.016 g H"_2/(1 color(red)(cancel(color(black)("mol H"_2)))) = "2.264 g H"_2`

(c). Molecules of `"NH"_3`

(i). Calculate the moles of `"H"_2`

`"Moles of H"_2 = 1.59 × 10^"-4" color(red)(cancel(color(black)("g H"_2))) × "1 mol H"_2/(2.016 color(red)(cancel(color(black)("g H"_2)))) = 7.887 × 10^"-5"color(white)(l) "mol H"_2`

(ii). Calculate the moles of `"NH"_3`

`" Moles of NH"_3 = 7.887 × 10^"-5" color(red)(cancel(color(black)("mol H"_2))) × "2 mol NH"_3/(3 color(red)(cancel(color(black)("mol H"_2)))) = 5.258 ×10^"-5"color(white)(l) "mol NH"_3`

(iii). Calculate the molecules of #"NH"_3

`"Molecules of NH"_3 = 5.258 ×10^"-5" color(red)(cancel(color(black)("mol NH"_3))) × (6.022 × 10^23 color(white)(l)"molecules NH"_3)/( 1 color(red)(cancel(color(black)("mol NH"_3)))) = 3.17 × 10^19color(white)(l)"molecules NH"_3`