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Stoichiometry – the calculation of quantities in chemical equations.

Stoichiometry – the calculation of quantities in chemical equations. When balancing a chemical equation, change only the coefficients, never change the subscripts. The coefficients tell how many molecules are in a reaction, the subscripts tell how many atoms are in a molecule.

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Stoichiometry – the calculation of quantities in chemical equations.

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  1. Stoichiometry – the calculation of quantities in chemical equations. When balancing a chemical equation, change only the coefficients, never change the subscripts. The coefficients tell how many molecules are in a reaction, the subscripts tell how many atoms are in a molecule.

  2. The coefficients are very important to stoichiometry-they tell the relationship between reactants and products in an equation. Things that are conserved in every reaction: • Energy • Mass • Number of atoms

  3. The coefficients can be: • A literal interpretation of how many actual molecules in an equation. 2 H2 + O2 2 H2O This means 2 molecules of hydrogen will react with 1 molecule of oxygen to give 2 molecules of water. This is good for balancing equations, but little else as molecules are too small to deal with in the lab.

  4. The coefficients can tell the number of moles of each molecule in the equation. 2 H2 + O2 2 H2O This means 2 moles of hydrogen reacts with 1 moles of oxygen will give 2 moles of water. • The coefficients can tell the number of volumes of gas in a reaction. This will work only if the molecules are gas molecules. 2 volumes of hydrogen reacts with 1 volume of oxygen will give 2 volumes of water.

  5. Moles Ratios – a fraction, in moles, of 2 different molecules in a balanced equation (a conversion factor for unit analysis). 2 H2 + O2 2 H2O There are three equivalencies (comparisons): 2 mol H2 ≈ 1 mol O2 2 mol H2 ≈ 2 mol H2O 1 mol O2 ≈ 2 mol H2O

  6. From each equivalency, two mole ratios can be formed: 2 mol H2/1 mol O2 1 mol O2/2 mol H2 2 mol H2/2 mol H2O 2 mol H2O/2 mol H2 1 mol O2/2 mol H2O 2 mol H2O/1 mol O2 Which mol ratio is used depends on which ratio is needed when doing unit analysis, or which one the question asks for.

  7. N2 + H2 NH3 • If you start with 3.0 moles of N2, how many moles of H2 are needed? How many moles of NH3 can be produced? • Balance the reaction: • N2 + 3 H2 2 NH3 • 2. Write out the equivalencies: • 1 mol N2≈3 mol H2; 1 mol N2≈2 mol NH3; 3 mol H2≈2 NH3 • 3. Set up the unit analysis

  8. 3.0 mol N2 3 mol H2= 9.0 mol H2 1 mol N2 3.0 mol N2 2 mol NH3 = 6.0 mol NH3 1 mol N2 The mole ratios are conversion factors using the coefficients from the balanced equation.

  9. The Double Flux Capacitor Part. A Mass A Part. B Mass A Mole A Moles B Vol. A Vol. B Particles  Moles 1 Mol = 6.022 E 23 things Mass  Moles 1 Mol = Molar Mass Volume  Moles 1 Mol = 22.4 L gas Moles A  Moles B coefficients from the balanced equation (mole ratios)

  10. What mass of I2 can be made from 10.0 liters of Cl2(g) according to the following reaction: Cl2 + 2 KI  2 KCl + I2 10.0 L Cl2 1 mol Cl2 1 mol I2 253.8 g I2 = 113 g I2 22.4 L Cl2 1 mol Cl2 1 mol I2

  11. How many grams of water can be made from 13.5 grams of H3PO4 according to the following reaction: • H3PO4 H4P2O7 + H2O • What volume of hydrogen gas is needed to make 15.3 grams of iron according to the reaction: • Fe2O3 + H2 Fe + H2O • 3. How many liters of CO2 can be made from 17.5 grams of oxygen: C20H42 + O2  CO2 + H2O • 4. How many molecules of Fe2O3 are needed to react with 18.5 liters of CO according to the reaction: • Fe2O3 + CO  Fe + CO2 • 5. How many grams of PbI4 can be made when 12.5 g CrI3react: PbO2+ CrI3  PbI4 + Cr2O3

  12. Theoretical Yield – the maximum amount of product that can be formed based upon a balanced equation (using the double flux capacitor). Actual Yield – the amount of product actually made in the lab. % Yield – (actual yield/theo. yield) x 100 % The % yield should always be less than 100 % due to experimental error. When plugging in the theoretical yield, always use the UNROUNDED value.

  13. The actual yield will always be one of the products. It can never be one of the reactants. Na + H2O  NaOH + H2 If 18.3 g of sodium can produce 8.23 liters of hydrogen gas, what is the % yield? Balanced equation: 2,2,2,1 18.3 g Na 1 mol Na 1 mol H2 22.4 l H2 = 8.91 l H2 23.0 g Na 2 mol Na 1 mol H2 (8.23 l H2 / 8.9113 … l H2) x 100 % = 92.4 %

  14. Limiting Reagent (Reactant) – the starting material that runs out first in a chemical reaction causing the reaction to stop. 2 B + 1 C + 3 H + 1 T + 3 L + 1 M  1 HS How many ham sandwiches can be made from 10 bread, 8 cheese, 12 ham, 7 tomato, 15 lettuce and 20 mustard? Bread  5 cheese  8 ham  4 Tomato  7 lettuce  5 mustard  20

  15. Ham is the limiting reactant (LR) because only 4 ham sandwiches can be made. After that point, the ham runs out causing the reaction to stop. Therefore no more sandwiches can be made. When working a limiting reactant problem, work the problem twice- calculate the amount of product that can be made from each reactant. The smallest amount of product will be how much that can be made. The reactant that causes the least amount of product will be the LR.

  16. N2 + 3 H2 2 NH3 What mass of NH3 can be made from 6.37 g of N2 and 10.3 g of H2? What is the LR? 6.37 g N2 1 mol N2 2 mol NH3 17.0 g NH3 = 7.74 g NH3 28.0 g N2 1 mol N2 1 mol NH3 10.3 g H2 1 mol H2 2 mol NH3 17.0 g NH3 = 58 g NH3 2.0 g H2 3 mol H2 1 mol NH3 7.74 grams of NH3 can be made in the reaction, and N2 is the LR.

  17. Stoichiometry • Only 1 given number – double flux capacitor problem. “theoretical yield” • 2 given numbers: 1 reactant and 1 product- double flux capacitor with the reactant going to same units as the product (theo yield). Product number is the actual yield. (actual/theo) x 100 % = % yield • 2 given numbers, both reactants. Limiting reagent problem. Double flux each starting number. Smallest product is the answer. Starting number that gives smallest product is LR.

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