00 mol h 2 o are produced the advantage of learning

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limiting reactant, and 2.00 mol H 2 O are produced. The advantage of learning this method is that it works even when the coefficients become difficult to use with the visual method. Example Aluminum chloride, AlCl 3 , has many uses including in deodorants and antiperspirants. It is synthesized from aluminum and chlorine. What mass of AlCl 3 can be produced if 100 g of each reactant are available? What is the limiting reactant? How many grams of the excess reactant remain? Step 1: Write the balanced equation. 2 Al(s) + 3 Cl 2 (g) → 2 AlCl 3 (s) Step 2: Set up the SRF Table. Since only moles can go into the table, the grams of each reactant will first need to be converted to moles. See Figure 24. 2  Al(s) + 3  Cl 2 (g) 2  AlCl 3 (s) S tarting moles 3.70 1.41 0 R eacting moles - 2 x - 3 x + 2 x F inal moles 3.70 - 2x 1.41 –3x 2x Step 3: Set the “Final moles” equal to zero and solve for x. Choose the smallest value of x. 3.70 – 2x = 0 or 1.41 – 3x = 0 x = 1.85 moles or x = 0.47 moles
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Since x = 0.47 mol is smaller, Cl 2 will be consumed first; it is the limiting reactant. Step 4: Determine the amount of product formed and the amount of Al remaining by substituting x = 0.47 into the corresponding equations on the “Final moles” line. Amount of product (AlCl 3 ): 2x = 2(0.47 mole) = 0.94 moles AlCl 3 Amount of Al remaining: 3.70 – 2(0.47 mole) = 2.76 moles of Al Step 5: Convert the moles back to grams.
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