1. How many grams of mercury (II) chloride can be prepared from 30.24 g of mercu

ID: 780374 • Letter: 1

Question

1. How many grams of mercury (II) chloride can be prepared from 30.24 g of mercury for the equations below?

a.) Hg(l) + 2H2S04 (aq) -----> HgSO4 (s) + @H2O (g) + SO2 (g)

b.) HgSO4 (s) + 2NaCl (s) ------> Na2SO4 (s) + HgCl2(s)

2. What is the theoretical (or calculated) yield of NH4Al(SO4)2 * 12H2O formed from 1.872g of Al(OH)3? Assume all other reagents are in excess.

a.) 2Al(OH)3 (s) + 3H2SO4 (aq) ------> Al2 (SO4)3 (aq) + 6H20 (l)

b.) Al2(SO4)3 (aq) + (NH4)2SO4 (aq) + 24H2O (l) ------> 2NH4Al(SO4)2 * 12H2O (s)

1) moles of Hg = 30.24/200.6 = 0.15075 ,

by stochiometry in eq we observe that 1 mole of Hg gives 1 mole of HgCl2 finally ,

hence HgCl2 moles = 0.15075 , HgCl2 mass = 0.15075 x 271.52 = 40.93 gm,

2) moles of Al(OH)3 = 1.872/78 = 0.024, moles of product gien = ( 0.024)

mass of product mentioned = ( 0.024 x 348.12) = 8.3554 gm

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