Question #740df

1 Answer
Aug 20, 2015

Answer:

You had 0.00190 moles of ammonia in the titration and 0.405 g of ammonia in the original sample of cloudy ammonia.

Explanation:

The idea behind this problem is that you need to work backward from the number of moles of hydrochloric acid used in the titration to determine how many moles of ammonia you get in the 20-mL aliquot.

Then use this value to determine how many noles of ammonia you'd get in the 250-mL sample.

So, ammonia and hydrochloric acid will react according to this balanced chemical equation

#"NH"_text(3(aq]) + "HCl"_text((aq]) -> "NH"_4"Cl"_text((aq])#

Since you have a #1:1# mole ratio between ammonia and hydrochloric acid, the number of moles of ammonia will be equal to the number of moles of hydrochloric acid used in the titration.

This means that you have

#C = n/V implies n = C * V#

#n_"HCl" = "0.0892 M" * 21.35 * 10^(-3)"L" = "0.001904 moles HCl"#

This of course means that the 20-mL aliquot contained

#0.001904color(red)(cancel(color(black)("moles HCl"))) * ("1 mole NH"""_3)/(1color(red)(cancel(color(black)("mole HCl")))) = "0.001904 moles NH"""_3#

You can use a simple proportion to figure out how many moles of ammonia were present in the 250-mL sample

#250color(red)(cancel(color(black)("mL"))) * ("0.001904 moles NH"""_3)/(20color(red)(cancel(color(black)("mL")))) = "0.0238 moles NH"""_3#

Since the 250-mL sample was prepared by diluting the original sample, the number of moles of ammonia present in the cloudy ammonia sample will be equal to the number of moles of ammonia present in the 250-mL sample.

The 250-mL sample was essentially prepared by adding water, the amount of ammonia was kept constant, which is what diluting a solution implies.

This means that the 23.27-g sample of cloudy ammonia contained 0.0238 moles of ammonia. Use ammonia's molar mass to determine how many grams would contain this much ammonia

#0.0238color(red)(cancel(color(black)("moles"))) * "17.03 g"/(1color(red)(cancel(color(black)("mole")))) = "0.405 g NH"""_3#

I'll leave the values rounded to three sig figs, despite the fact that you were very inconsistent with the number of sig figs provided

#n_("NH"""_3"titration") = color(green)("0.00190 moles")#

and

#m_("NH"""_3) = color(green)("0.405 g")#