Question

Be sure to answer all parts. A 0.9340 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 1.913 g of AgCl. Calculate the percent by mass of each compound in the mixture.

% mass NaCl

% mass KCl

Answer 1

Lets take mass of Nach m(Nad) = x H & mass of kcl mckelly mp_ Wall & Ag No → Agd + NaNO₃ Number of - Number of moles of Nae moles of Agcl Let mass of Ag Cl be m e number of moles mass molar mass molar mass of Nach molar mass of Age Mis 143.322 - 58.44 and is m = 2.450 (9) ka & AgNO₃ → Agd + KNO Number of Number of moles of kl males of Ag of let mass of Ag d here be my ya ma 74.55 143.32 M 143.324 74.55 L m₂ = 1.92 y (g) Now, as per given data - M, + m2 = 1:913 (9) 2:45% + 1924 = 1.913. 0 Also, m (Nad) + mckel) = 0.9340g 2 + y = 0.9340_-(2) 09340-4 put this in (1) 2.45 10.9340-4) 1.92y = 1.913 0

Data 2:2283 - 2:45y + 1.924 = 1.913 2.2.883 -1.913 = 2.454 -1.92y 0:3753 = 0.53y y = 0.3753 /0.53 y 20.7081 x= 0.9 340-y x = 0.9340 - 07081 x = 0.2259 lets calculate to Wall& 1 ku- Ora Nad = 0.22599 Kul= 0.70819 - total = 0.93409 f. Nad = mass of Nach x 100 total mass 0.2259 g x 100 0.9340 = 24.19%. f.kd = 0.70819 x 100 0.93409 = 75.81 A