# Question d8ba0

Jun 18, 2017

The suspension contains 5.32 % ammonia by mass.

#### Explanation:

Write the balanced chemical equation for the neutralization

${M}_{\textrm{r}} : \textcolor{w h i t e}{l} 17.03$
$\textcolor{w h i t e}{m m m} \text{NH"_3 + "HCl" → "NH"_4"Cl}$

Calculate the mass of ${\text{NH}}_{3}$ in the titration aliquot

$\text{Moles of HCl" = 0.0256 color(red)(cancel(color(black)("L HCl"))) × "0.0924 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = 2.365 × 10^"-3"color(white)(l) "mol HCl}$

${\text{Moles of NH"_3 = 2.365 × 10^"-3"color(red)(cancel(color(black)("mol HCl"))) × "1 mol NH"_3/(1 color(red)(cancel(color(black)("mol HCl")))) = 2.365 × 10^"-3"color(white)(l) "mol NH}}_{3}$

${\text{Mass of NH"_3 = 2.365 × 10^"-3" color(red)(cancel(color(black)("mol NH"_3))) × "17.03 g NH"_3/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "0.040 28 g NH}}_{3}$

Calculate the mass of ${\text{NH}}_{3}$ in the original sample

20 mL of solution contain 0.040 28 g ${\text{NH}}_{3}$

∴ In 250 mL of solution,

${\text{Mass of NH"_3 = "0.040 28"color(white)(l) "g NH"_3 × (250 color(red)(cancel(color(black)("mL"))))/(20 color(red)(cancel(color(black)("mL")))) = "0.5035 g NH}}_{3}$

Calculate the % by mass of ${\text{NH}}_{3}$

"% by mass" = (0.5035 color(red)(cancel(color(black)("g"))))/(9.46 color(red)(cancel(color(black)("g")))) × 100 % = 5.32 %#