18. Calculate the percent yield of alum if you started the experiment with 0.53g

Question

18. Calculate the percent yield of alum if you started the experiment with 0.53g of Al and ended with 10.42g of wet Alum. The equation for the synthesis of alum is below. 2 Al(s) + 2 KOH (aq) + 6 H2O(l) 3H2(g) + 2 KAI(OH)4 (aq) KAl(OH)4(a) 2 H2SO4 a) +8 H2O(1) KAl(SO4)2 12 H20(aq) (a) 89% (b) 12% (c) 10% (d) 112% 19. Which of the following molecules does not exhibit a molecular dipole moment? (a) SCl2 (b) NH3 (c) NF3 (d) BF3 20. Creating the organic compound luminol is a two-step process and requires isolating luminol and the intermediate. What laboratory technique was used to isolate the two compounds? (a) filtration (b) heating (c) separating funnel (d) crystallization

Explanation / Answer

18) alum is prepared from aluminium

As per balanced equation, 2 moles of Al will give two moles of KAl(OH)4

Each mole of KAL(OH)4 will give one mole of alum

so in all one mole of aluminium will give 1 mole of alum

the mass of aluminium used = 0.53 g

Atomic weight of aluminium = 27 g / mole

Moles of aluminium = mass / at wt = 0.53 / 27 g / mole = 0.0196 moles

so theoretical moles of alum that should be formed = 0.0196

Theoreticl mass of alum obtained = Moles X molar mass =0.0196 X 474 = 9.290g

Mass of alum obtained = 10.42g

so % yeild = 10.42 X 100 / 9.290 = 112.1%

19) BF3 is non polar and does not exhibit a molecular dipole moment.

The B-F bond is polar but the dipoles cancel each other and hence net dipole is zero

20) In this process we will use filtration.

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