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Have out for today

Have out for today. Your periodic table Your scientific calculator. 9.1 Mole Ratios. Now that we know what a balanced equation means, we can use an equation to predict the moles of products that a given number of moles of reactants will yield. For example: 2H 2 0(l)  2H 2 (g) + O 2 (g)

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Have out for today

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  1. Have out for today • Your periodic table • Your scientific calculator

  2. 9.1 Mole Ratios • Now that we know what a balanced equation means, we can use an equation to predict the moles of products that a given number of moles of reactants will yield. • For example: 2H20(l)  2H2(g) + O2(g) 2 mol H2O yields 2 mol H2 and 1 mol O2

  3. So what if… I start with… 2H20(l)  2H2(g) + O2(g) Then instead of 2 mol, I have 4 mol of H2O…how many mol of each product do I get? 4H20(l)  __H2(g) + __O2(g) 4 2

  4. Now suppose… Given the same equation 2H20(l)  2H2(g) + O2(g) If we decompose 5.8 mol of water, how many moles of products are formed? Start with O2. Let’s use a mole ratio – ratio of moles of one substance to moles of another substance in a balanced chemical equation

  5. and 2H20(l)  2H2(g) + O2(g) 2 mol H2O = 1 mol O2 forms 2 conversions factors: Start with the number given in the problem

  6. And since… 2H20(l)  2H2(g) + O2(g) …the ratio of water to hydrogen is the amount of H2 produced is 5.8 mol.

  7. So try this. • Calculate the number of moles of oxygen required to react exactly with 4.3 mol of propane, C3H8, in the reaction described by the following balanced equation. C3H8 (g) + 5O2(g) 3CO2(g) + 4H2O(g) 4.3 mol C3H8 ?? mol O2

  8. Last example • Ammonia is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation: N2(g) + 3H2(g) 2NH3(g) Calculate the moles of ammonia that are produced from 1.3 mol of hydrogen reacting with excess N2

  9. One Step Stoichiometry Worksheet

  10. 9.2 Mass calculations Consider the reaction of powdered Al and finely ground iodine to produce aluminum iodide. Al(s) + I2(s)  AlI3(s) What mass of iodine is needed to react with 35.0 g solid aluminum? FIRST: be sure the equation is balanced. 2Al(s) + 3I2(s)  2AlI3(s)

  11. Because the balanced equation deals with moles instead of grams • SECOND – convert 35.0 grams of aluminum to mol of aluminum THIRD – Use the coefficients to determine the mole ratio required to go from mol of Al to mol I2 and convert. Mole ratio

  12. Last step! FOURTH – Convert the number of moles of I2 into grams. So…given 2Al(s) + 3I2(s)  2AlI3(s) It takes 495g of I2 to react with 35.0g of Al

  13. Now, given the same reaction… 2Al(s) + 3I2(s)  2AlI3(s) Find the mass of AlI3(s) formed by the 35g of Al(s) with 495g of I2(s).

  14. Homework • Section 9.2 review • # 2,4 & 6 only! • Due tomorrow!!! I will not accept this late for credit!

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