(a) When 1.50 mol of Al and 3.00 mol of Cl2 combine in reaction 2 Al(s) + 3 Cl2(

Question

(a) When 1.50 mol of Al and 3.00 mol of Cl2 combine in reaction 2 Al(s) + 3 Cl2(g) -----> 2 AlCl3(s), which is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the excess reactant remain at the end of the reaction?

Explanation / Answer

Sorry in the previous answer by mistake I took value of Cl for Al and value of Al for Cl. CORRECT ANSWER.... 2 Al(s) + 3 Cl2(g) -----> 2 AlCl3(s) a).Moles Al = 1.50 Moles Cl2 = 3.00 the ratio is 1:2 Al is the limiting reactant. b).the ratio between Cl2 and AlCl3 is 2 : 2 moles AlCl3 = 3.00 x 2 / 2 = 1.50 moles Mass AlCl3 = 1.50 mol x 133.3 g/mol= 199.5 g c).We have 1.50 moles of Al and 3 mole of Cl2 so we can calculate mass of exess of Cl2 which comes out to be 1.50 mol Cl2 i.e. 0.75 mol Cl

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