2HCl + Ca(OH)2 → CaCl2 + 2H2O We don’t have enough calcium hydroxide

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Presentation transcript:

2HCl + Ca(OH)2 → CaCl2 + 2H2O We don’t have enough calcium hydroxide b) g HCl mol HCl mol Ca(OH)2 needed g Ca(OH)2 needed 1 mol HCl 1 mol Ca(OH)2 74.093 g Ca(OH)2 1 mol Ca(OH)2 x 230.00 g HCl x x = 233.7g Ca(OH)2 needed 36.46 g HCl 2 mol HCl Given: 210.00 g Ca(OH)2 We don’t have enough calcium hydroxide Thus, Ca(OH)2 is the limiting reagent

c) HCl d) Use limiting reagent, Ca(OH)2, to calculate amount of HCl that reacts with it. g Ca(OH)2 mol Ca(OH)2 mol HCl reacted g HCl reacted 36.46 g HCl 1 mol Ca(OH)2 74.093 g Ca(OH)2 x 2 mol HCl 1 mol Ca(OH)2 x 210.00 g Ca(OH)2 x = 1 mol HCl 206.68 g HCl reacted Thus . . . . . . . . starting with 230.00 g HCl - 206.68 g HCl reacted 23.32 g HCl in excess

Complete the other three reactions for next class g Ca(OH)2 mol Ca(OH)2 mol CaCl2 g CaCl2 1 mol Ca(OH)2 74.093 g Ca(OH)2 x 1 mol CaCl2 1 mol Ca(OH)2 x 110.98 gCaCl2 1 mol CaCl2 x = 210.00 g Ca(OH)2 314.55 g CaCl2 314.55 g CaCl2 % Yield = Actual Yield Theoretical Yield x 100 f) 294.62 % Yield = x 100 93.7% = 314.55 Complete the other three reactions for next class