hi, I\'m stuck in this lab those are the data from the lab 1-mass of weighing pa

Question

hi, I'm stuck in this lab

those are the data from the lab

1-mass of weighing paper for alum (g) = 3.9123

2-mass of weighing paper plus alum sample (g) = 4.7238

3-mass of empty heated and cooled crucible (g) = 25.7774

4-mass of crucible plus BaSO4 after first heating (g) = 25.8974

5-mass of crucible plus BaSO4 after second heating (g) = 25.8972

Find the calculation for :

1- Mass of Alum (g) = 0.8115 (g)

2- Mass of BaSO4 (g) = 0.12 (g)

3- Percent SO4^2- in alum sample ?

4-Theoretical percent SO4^2- in the pure alum sample ?

5-Analytical purity of the alum sample?

and this is the formela tha I came up with

KAl(SO4)2*2H2O(aq) + 2 BaCl2(aq) --> 2 BaSo4(s) + KCl(aq) + AlCl3(aw) + 12 H2O(l)

Explanation / Answer

Mass of alum = mass of weighing paper plus alum sample - mass of weighing paper for alum

mass of alum = 4.7238 - 3.9123

mass of alum = 0.8115

Mass of BaS04 = mass of crucible plus BaSO4 after first heating - mass of empty heated and cooled crucible

mass of BaS04 = 25.8974 - 25.7774

mass of BaS04 = 0.12 g

moles of BaS04 = 0.12 /233.43

moles of BaS04 = 5.14 x 10-4

total moles of Baso4 = 5.14 x 10-4 x 2

total moles of BaS04 = 10.28 x 10-4

moles of S042- = 10.28 x 10-4

mass of So42- = 10.28 x 10-4 x 96 = 0.098688 grams

these mass has come from the alum

percent of So4 2- in alum = 0.098688 x 100 / 0.8115 = 12.16 %

theroetical percentage = mass of So4 in alum x 100 / Molar mass of alum

therotical = 192 x 100/474

therorectical = 40.5 %

percentage purity = (12.16 / 40.5 ) x 100

percentage purity = 30.02

purity of alum sample is 30.02 %

the reaction is given by

KAl(SO4)2*12H2O(aq) + 2 BaCl2(aq) --> 2 BaSo4(s) + KCl(aq) + AlCl3(aw) + 12 H2O(l)

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