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The BCA Method in Stoichiometry

The BCA Method in Stoichiometry. Step 1- Balance the equation.

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The BCA Method in Stoichiometry

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  1. The BCA Method in Stoichiometry

  2. Step 1- Balance the equation Hydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur dioxide and water. How many moles of oxygen gas would be needed to completely burn 2.4 moles of hydrogen sulfide? 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before:Change After

  3. Step 2: fill in the before line 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before: 2.4 xs 0 0Change After • Assume more than enough O2 to react

  4. Step 3: use ratio of coefficients to determine change 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4 After • Reactants are consumed (-), products accumulate (+)

  5. Step 4: Complete the table 2 H2S + 3 O2 ----> 2 SO2 + 2 H2O Before: 2.4 xs 0 0Change –2.4 –3.6 +2.4 +2.4After 0 xs 2.4 2.4

  6. Only moles go in the BCA table • The balanced equation deals with how many, not how much. • If given mass of reactants for products, convert to moles first, then use the table.

  7. Limiting reactant problems • Distinguish between what you start with and what reacts. • When 0.50 mole of aluminum reacts with 0.72 mole of iodine to form aluminum iodide, how many moles of the excess reactant will remain?How many moles of aluminum iodide will be formed? 2 Al + 3 I2  2 AlI3 Before: 0.50 0.72 0Change After

  8. Limiting reactant problems • Guess which reactant is used up first, then check 2 Al + 3 I2  2 AlI3 Before: 0.50 0.72 0Change -0.50 -0.75 After • It’s clear that there’s not enough I2 to react with all the Al.

  9. Limiting reactant problems • Now that you have determined the limiting reactant, complete the table, then solve for the desired answer. 2 Al + 3 I2  2 AlI3 Before: 0.50 0.72 0Change -0.48 -0.72 +0.48 After 0.02 0 0.48

  10. Complete calculations on the side • In this case, desired answer is in moles • If mass is required, convert moles to grams in the usual way • 3.6 moles O2 x 32.0g = 115 g O2 1 mole

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