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Chemical Equations IV

Chemical Equations IV. Stoichiometric Calculations. Coefficent in Balanced Equation. A. Gives the relative number of molecules of reactants and products. Example: N 2(g) + 3H 2(g) --- > 2NH 3(g)

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Chemical Equations IV

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  1. Chemical Equations IV Stoichiometric Calculations

  2. Coefficent in Balanced Equation A. Gives the relative number of molecules of reactants and products. Example: N2(g) + 3H2(g) --- > 2NH3(g) One molecule of nitrogen gas molecules plus 3 molecules of hydrogen gas molecules forms 2 molecules of ammonia gas.

  3. Coefficent in Balanced Equation • Gives the relative number of moles of reactants and products. Example: N2(g) + 3H2(g) --- > 2NH3(g) One mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas

  4. Law of Conservation of Mass Mass of reactants = Mass of product Example: N2(g) + 3H2(g) --- > 2NH3(g) 2(14.0 g/mol) + 6(1.01 g/mol) = 2(14.0g/mol) + 6(1.01 g/mol) 34.06 g/mol = 34.06 g/mol

  5. Steps to a Correct Answer • Write a balanced equation. • Write a dimensional analysis set-up with labeled values. Start with given value in problem. 3. Don’t do work in parts. I want to see one step. 4. Calculate answer

  6. Type: Mole-Mole moles of ---> moles of reactant product Use coefficients in balanced equation Put moles of chemical you’re solving for in the numerator. Stoichiometry I

  7. Example 1. Mole-Molestart with mole – end with mole How many moles of nitric oxide (nitrogen monoxide) are produced when 0.500 moles of nitrogen reacts with excess nitrogen? Step 1: N2(g) + O2(g) --- > 2NO(g) Step 2: Start with the given value. Write mole ratio using coefficients in balanced equation. Put the number of moles of the substance you’re solving for in the numerator. 0.500 mol N2 x 2 mol NO 1 mol N2 Step 3: Answer: 1.00 mol NO

  8. Type: Mass – Mole mass of ---> moles of reactant product Change mass of chemical given to moles using it’s molar mass. Mole ratio from balanced equation. Put moles of chemical you’re solving for in numerator. Stoichiometry II

  9. Example 2. Mass-Molestart with mass – end with mole C2H5OH(ℓ) + 3O2(g) ---- > 2CO2(g) + 3H2O(ℓ) How many grams of CO2 are produced when 3.00 g of C2H5OH is burned? 3.00 g C2H5OH X 1 mol C2H5OH X 2 mol CO2 46.08 g C2H5OH 1 mol C2H5OH Answer: 0.130 mol CO2

  10. Type: Mass – Mass mass of ---> mass of reactant product Change mass to moles using molar mass Use mole ratio from balanced equation. Put moles of chemical you’re solving for in the numerator. Change moles to mass using molar mass Stoichiometry III

  11. Example 3. Mass – Massstart with mass – end with mass 2 C8H18(ℓ) + 25 O2(g) ---- > 16 CO2(g) + 18 H2O(ℓ) How many grams of O2 are needed to burn 2.00 g of C8H18? 2.00 g C8H18 X 1 mol C8H18X 25 mol O2 x 32.0 g O2 114.26 g C8H18 2 mol C8H18 1 mol O2 Answer: 7.00 g O2

  12. Type: Moles – Mass moles of ---> mass of reactant product Change mass of given chemical to moles using molar mass Use mole ratio from balanced equation putting moles of chemical you’re solving for in numerator. Change moles to mass using molar mass Stoichiometry IV

  13. Example 4. Moles to Massstart with moles – end with mass 2 C4H10(ℓ) + 13 O2(g) ---- > 8 CO2(g) + 10 H2O(ℓ) How many grams of CO2 are produced when 1.72 x 10-2 moles of C4H10 burns in air? 1.72 x 10-2 mol C4H10 X 8 mol CO2 X 44.01 g CO2 2 mol C4H10 1 mol CO2 Answer: 3.03 g CO2

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