Let's combine these four equations into one big equation. We will have to multiply the first two equations by 2 to make things cancel.
#"2NH"_3 +stackrelcolor(blue)(1)(color(red)(cancel(color(black)(2))))"H"_2"O" + stackrelcolor(blue)(1)(color(red)(cancel(color(black)(2))))"CO"_2 → color(red)(cancel(color(black)("2NH"_4"HCO"_3)))#
#color(red)(cancel(color(black)("2NH"_4"HCO"_3))) + "2NaCl" → "2NH"_4"Cl" + color(red)(cancel(color(black)("2NaHCO"_3)))#
#color(red)(cancel(color(black)("2NaHCO"_3))) → color(red)(cancel(color(black)("Na"_2"CO"_3))) + color(red)(cancel(color(black)("H"_2"O"))) + color(red)(cancel(color(black)("CO"_2)))#
#color(red)(cancel(color(black)("Na"_2"CO"_3))) + "10H"_2"O" → "Na"_2"CO"_3·"10H"_2"O"#
#stackrel(————————————————————)("2NH"_3 + "CO"_2 + "2NaCl" + "11H"_2"O" → "2NH"_4"Cl" + "Na"_2"CO"_3·10"H"_2"O")#
Calculate the mass of #"NH"_3#
#"Moles of Na"_2"CO"_3·10"H"_2"O"#
#= 200 color(red)(cancel(color(black)("kg Na"_2"CO"_3·10"H"_2"O"))) × (1 color(red)(cancel(color(black)("kmol Na"_2"CO"_3·10"H"_2"O"))))/(286.14 color(red)(cancel(color(black)("kg Na"_2"CO"_3·10"H"_2"O")))) = "0.6990 kmol Na"_2"CO"_3·10"H"_2"O"#
#"Moles of NH"_3#
#= 0.6990 color(red)(cancel(color(black)("kmol Na"_2"CO"_3·10"H"_2"O"))) × (2 color(red)(cancel(color(black)("kmol NH"_3))))/(1 color(red)(cancel(color(black)("kmol Na"_2"CO"_3·10"H"_2"O")))) = "1.398 kmol NH"_3#
#"Mass of NH"_3 = 1.398 color(red)(cancel(color(black)("kmol NH"_3))) × ("17.03 kg NH"_3)/(1 color(red)(cancel(color(black)("kmol NH"_3)))) = "23.8 kg NH"_3#
Calculate the mass of #"NaCl"#
#"Moles of NaCl"#
#= 0.6990 color(red)(cancel(color(black)("kmol Na"_2"CO"_3·10"H"_2"O"))) × (2 color(red)(cancel(color(black)("kmol NaCl"))))/(1 color(red)(cancel(color(black)("kmol Na"_2"CO"_3·10"H"_2"O")))) = "1.398 kmol NaCl"#
#"Mass of NaCl" = 1.398 color(red)(cancel(color(black)("kmol NaCl"))) × ("58.44 kg NaCl")/(1 color(red)(cancel(color(black)("kmol NaCl")))) = "81.7 kg NaCl"#
