Stoichiometry Chemical Quantities Chapter 9. What is stoichiometry? stoichiometry- method of determining the amounts of reactants needed to create a certain.

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Presentation transcript:

Stoichiometry Chemical Quantities Chapter 9

What is stoichiometry? stoichiometry- method of determining the amounts of reactants needed to create a certain amount of products. we will combine balancing equations with molar mass and add one more step.

How to do it the coefficients in a balanced equation are ratios of the molecules needed for the reaction. moles are just a number of molecules so2 H 2 + O 2  2 H 2 O How many moles of H 2 O are made by completely reacting 1.4 mol of O mol O 2 2 mol H 2 O 1 mol O 2 =2.8 mol H 2 O

Adding in molar mass How many grams of H 2 are needed to completely react with 24 g of O 2 ? 2 H 2 + O 2  2 H 2 O convert to moles  convert to the desired molecule  convert to grams O 2 = g/mol ; H 2 = g/mol 24 g O 2 1 mol O 2 2 mol H g H 2 32 g O 2 1 mol O 2 1 mol H 2 = 3.0 g H 2

Practice How many grams of H 2 O are produced from 15 g of H 2 ? 2 H 2 + O 2  2 H 2 O 15 g H 2 1 mol H 2 2 mol H 2 O g H 2 O g H 2 2 mol H 2 1 mol H 2 O = 130 g H 2 O

practice How many grams of H 2 O are produced by completely reacting 24 g of O 2 ? 2 H 2 + O 2  2 H 2 O convert to moles  convert to the desired molecule  convert to grams O 2 = g/mol ; H 2 O = g/mol 24 g O 2 1 mol O 2 2 mol H 2 O g H 2 O 32 g O 2 1 mol O 2 1 mol H 2 O = 27 g H 2 O

Another How many grams of H 2 are needed to form 34 g of H 2 O? 2 H 2 + O 2  2 H 2 O convert to moles  convert to the desired molecule  convert to grams H 2 = g/mol ; H 2 O = g/mol 34 g H 2 O1 mol H 2 O2 mol H g H g H 2 O 2 mol H 2 O1 mol H 2 = 3.8 g H 2

Balance the equation 1 st, then do the problem Na + Cl 2  NaCl How many grams of chlorine are needed to react with 37 g of Na? g Na1 mol Na1 mol Cl g Cl g Na2 mol Na1 mol Cl 2 =57 g Cl 2

More CH 4 + O 2  H 2 O + CO 2 How many grams of CH 4 are needed to make 120 g of H 2 O? g H 2 O1 mol H 2 O1 mol CH g CH g H 2 O 2 mol H 2 O1 mol CH 4 = 53 g CH 4