C 15 Limiting Reagents PDF

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intro to limiting reagents and formulas...

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Some Stoichiometry H2(g) + Cl2(g) à2 HCl(g) 10.0 grams each of H2 and Cl2

10.0 g H2 (1 molH2) (2 molHCl) (36.46 g ) = 362 g HCl ----------------------------------------(2.016 g) (1 molH2) (1 molHCl) 10.0 g Cl2 (1 molCl2) (2 molHCl) (36.46 g ) = 10.3 g HCl ------------------------------------------(70.90 g ) (1 molCl2) (1 molHCl) Limiting Reagent - The limiting reagent in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. - The limiting reagent is HCl because it runs out first. Example: Solid Manganese (Mn) reacts with bromide gas to form solid manganese (III) bromide. 2 Mn(s) + 3Br2(g) ---> 2 MnBr3(s) - Each substance is 0.25 mol of a substance

9 X 0.25 mol Br2 = 2.25 mol Br2 2.25 mol Br2 (2mol MnBr3) ---------------- = 1.50 mol MnBr3 (3 mol Br2)

8 X 0.25 mol Mn = 2.00 mol Mn 2.00 mol Mn (2 mol MnBr3) ----------------- = 2.00 mol MnBr3 (2 mol Mn) *The Br2 will run out first. It is the limiting reagent. 1.50 mol MnBr3 is the theoretical yield. * Some Mn will be leftover and still present, how much?

Calculate the Excess Mn 2 Mn (s) + 3Br2 (g) ----> 2MnBr3 (s) 1) Use the limiting reagent to calculate how much excess reagent is used up 2.25 mol Br2 (2mol Mn) = 1.50 mol Mn is used up ------------(3 mol Br2) 2) By subtraction, calculate how much is leftover 2.00 mol Mn - 1.50 mol Mn used up = 0.50 mol Mn leftover...