Write the balanced chemical equation for the neutralization
#M_text(r):color(white)(l)17.03#
#color(white)(mmm)"NH"_3 + "HCl" → "NH"_4"Cl"#
Calculate the mass of #"NH"_3# in the titration aliquot
#"Moles of HCl" = 0.0256 color(red)(cancel(color(black)("L HCl"))) × "0.0924 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = 2.365 × 10^"-3"color(white)(l) "mol HCl"#
#"Moles of NH"_3 = 2.365 × 10^"-3"color(red)(cancel(color(black)("mol HCl"))) × "1 mol NH"_3/(1 color(red)(cancel(color(black)("mol HCl")))) = 2.365 × 10^"-3"color(white)(l) "mol NH"_3#
#"Mass of NH"_3 = 2.365 × 10^"-3" color(red)(cancel(color(black)("mol NH"_3))) × "17.03 g NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "0.040 28 g NH"_3#
Calculate the mass of #"NH"_3# in the original sample
20 mL of solution contain 0.040 28 g #"NH"_3#
∴ In 250 mL of solution,
#"Mass of NH"_3 = "0.040 28"color(white)(l) "g NH"_3 × (250 color(red)(cancel(color(black)("mL"))))/(20 color(red)(cancel(color(black)("mL")))) = "0.5035 g NH"_3#
Calculate the % by mass of #"NH"_3#
#"% by mass" = (0.5035 color(red)(cancel(color(black)("g"))))/(9.46 color(red)(cancel(color(black)("g")))) × 100 % = 5.32 %#
