A sample of aluminum sulfate, Al_2(SO_1)_3, (MM = 342.17 g/mol) has a mass of 78

Question

A sample of aluminum sulfate, Al_2(SO_1)_3, (MM = 342.17 g/mol) has a mass of 78.452 g. Calculate (i)the number of moles of the compound, (ii) the number of moles of S atoms, and (ii) the number of moles of O atoms. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye. In the following reaction, 50.0 mL, of 0.115 M Pb(NO_3)_2 solution reacts with a solution containing 1.57 g of NaI (MM = 149.89 g/mol). Pb(NO_3)_2(aq) + 2NaI(aq) rightarrow 2 NaNO_3(aq) + PbI_2(s) a) Calculate the number of NO_3^- ions in the Pb(NO_3)_2 solution. b) Determine the limiting reactant. c) Which reactant is the excess reactant and how many moles of it is left over at the end of the reaction. d) Calculate the theoretical yield of Pbl_2(MM = 461 g/mol).

Explanation / Answer

1) (i)moles of Aluminium sulfate [Al2(SO4)3] = [mass of Aluminium sulfate] / [molar mass] = 78.452 gm / 342.17 g/mole = 0.23 mole

(ii) one mole of Aluminium sulfate contains three moles of sulfur

thus,  0.23 mole of Aluminium sulfate contains = 3 x 0.23 = 0.69 mole Sulfur

(iii) one mole of Aluminium sulfate contains 12 moles of Oxygen

thus, 0.23 mole of Aluminium sulfate contains = 12 x 0.23 = 2.76 mole Oxygen

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