Stoichiometry!

Homework Page 285, #1-3 Review P. 288 – 295 P. 295, # 1-4 due Monday For tomorrow!

Stoichiometry! Composition stoichiometry – Reaction stoichiometry –

Mole Ratios In a nutshell: How many moles of one (product or reactant) will you need/receive if you are given one amount of another? (confused? Let’s do an example!) What is the molar ratio between magnesium and hydrochloric acid?

Mole Ratios What are all possible mole ratios for the following? How might these be useful?

Stoichiometry! Ratios of moles in a chemical equation can be used as conversion factors In the above reaction, how many grams of magnesium will you need to react with 2.00 moles of hydrochloric acid?

Practice! Determine all of the possible mole ratios for each of the following unbalanced reactions. ___Ba(ClO3)2  ___BaCl2 + ___O2 ___Ca + ___H2O  ___Ca(OH)2 + ___H2 ___MgCl2  ___Mg + ___Cl2 ___ Fe + ___Cu(NO3)2  ___Fe(NO3)2 + ___Cu ___KOH + ___H2SO4  ___K2SO4 + ___H2O ___Mg + ___O2  ___MgO ___AgNO3 + ___NaCl  ___AgCl + ___NaNO3 ___C2H6 + ___O2  ___CO2 + ___H2O

Stoichiometry These should be set up as unit conversions! These types are extremely important for determining the amount of product you should have at the end of an experiment Let’s practice!

Practice You have 13.82 grams of iron (II) sulfide, which you use as a reactant in the following formula equation: FeS + HCl  FeCl2 + H2S How much hydrochloric acid is needed in this reaction? Give your answer in grams. How much of each product is produced (in grams)? When methane combusts, there are 92.32 grams of oxygen gas available. How much methane is consumed in the reaction? How much of each product is produced (in grams)?

Homework P.295, #1-5 – due in three school days

Limiting Reactants Limiting reactant – Excess reagent – Limiting reactants occur whenever you do NOT have an ideal stoichiometric calculation.

Limiting Reactants Step 1 – convert everything to moles Step 2 – balance formula equation Step 3 – compare molar values to the balanced chemical equations Step 4 – determine the limiting reactant! Step 5 – calculate the amount of the second reactant needed Step 6 – subtract the amount of excess reactant needed from the excess amount available (this is your excess!) Step 7 – determine the amount of products produced based on your limiting reactant (use stoichiometry!)

Limiting Reactant - example Silicon dioxide (quartz) is usually quite unreactive, but readily reacts with hydrogen fluoride according to the following equation. SiO2 (s) + HF(g)  SiF4 (g) +2H2O (l) If 6.0 moles of hydrogen fluoride are added to 4.5 moles of silicon dioxide, which is the limiting reactant?

More practice – class work Some rocket engines use a mixture of hydrazine, N2H4, and hydrogen peroxide, H2O2, as the propellant. The formula equation is given by the following equation. N2H4(l) + H2O2(l)  N2(g) + 4H2O (g) Which is the limiting reactant in this reaction, when 0.750 moles of N2H4 is mixed with 0.500 moles of hydrogen peroxide? How much of the excess reactant, in moles, remains unchanged? How much of each product, in moles, is formed?

More Practice! The black oxide of iron, Fe3O4, occurs in nature as the mineral magnetite. This substance can also be made in the laboratory by the reaction between red-hot iron and steam according to the following equation. Fe(s) + H2O (g)  Fe3O4 (s) +H2 (g) When 36.0 g of steam are mixed with 67.0 grams of iron, which is the limiting reactant? What mass in grams of black iron oxide is produced? What mass in grams of excess reactant remains when the reaction is completed?

Percent Yield Percent Yield Example: Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disenfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation. C6H6(l) + Cl2 (g)  C6H5Cl (l) + HCl (g) When 36.8 grams of benzene reacts with excess chlorine gas, the actual yield of chlorobenzene is 38.8 grams. What is the percentage yield of chlorobenzene? Step 1: balance equation Step 2: Determine molecular weights Step 3: Determine theoretical yield Step 4: Determine percent yield = 73.2 %

Practice! #3 on page 302