Answer Key to Stoichiometric Problems Based on Chapter 3,4 of Kotz (Complete)

1. How many significant figures are there in 4.205 (4), 0.0056320 (5) , 146000 (3), 6.022x10²³ (4) ?
2. Convert these numbers to scientific notation: 4578 = 4.578 x 10³, 234500 = 2.345 x 10⁵, 0.0351 = 3.51 x 10^-2, 0.00021608 = 2.1608 x 10^-4
3. If 100.0 g of Al₂O₃ react with 100.0 g of H₂SO₄, how many grams of Al₂(SO₄)₃ are produced?
   Al₂O₃ + 3 H₂SO₄ -> Al₂(SO₄)₃ + 3 H₂O Answer = 116.3 g.
   The H₂SO₄ was limiting.
4. 2 C₅H₁₂O + 15 O₂ -> 12 H₂O + 10 CO₂
5. 2 Al + 6 HCl -> 2 AlCl₃ + 3 H₂
6. The 75.00 g of aluminum metal can produce 370.6 g of aluminum chloride. The 300.0 g of HCl can produce 365.7 g of aluminum chloride. The HCl is limiting and the answer = 365.7 g
   Al: 75.00 g of Al * (1 mol Al / 26.98 g) * ( 2 mol AlCl₃ / 2 mol Al) * (133.34 g / 1 mol AlCl₃ = 370.6 g
   HCl: 300.0 g * ( 1 mol / 36.34 g) * ( 2 mol AlCl₃ / 6 mol HCl) * (133.34 g / 1 mol AlCl₃ = 365.7 g
7. The 15.37 g of aluminum metal can produce 0.8545 mol of H₂. P = 745/760 = 0.9803 atm. T = 293 K. V = 21.0 L (only 3 sig fig allowed).
   mol of H₂: 15.37 g of Al * (1 mol Al / 26.98 g) * (3 mol H₂ / 2 mol Al) = 0.8545 mol
   Vol of H₂ = V = n R T / P = 0.8545 mol * (0.082057 L atm / mol K) * 293 K / 0.9803 atm = 21.0 L
8. For the compound C₃H₇Cl the mole weight is 78.54 g. C = 45.88%, H = 8.98%, and Cl = 45.14%. Note how the one heavy Cl atom is a large % of the total.
9. Another organic compound is 47.60% carbon, 7.21% hydrogen, and 45.19% fluorine by weight. Calculate the simplest empirical formula, C_xH_yF_z for the compound.
   Assume exactly 100 g of the compound, so there are 47.60 g of C, 7.21 g of H, 45.19 g of F.
   Determine moles of each: C = 47.60/12.01 = 3.963, H = 7.21/1.008 = 7.153, F = 45.19/18.998 = 2.379.
   Now divide by the smallest number of moles (F) to get ratios of moles as integers.
   C = 3.963/2.379 = 1.666, H = 7.153/2.379 = 3.007, F = 2.379/2.379 = 1.
   The value for H is close enough to assume it is 3, but the value for C = 1.666 is not close enough
   to 2 to assume it is that value. One should only round values which are very close to whole numbers.
   Since 1/3 = 0.333 and 2/3 = 0.667 try multiplying by 3.
   C = 3 * 1.666 = 4.998 = 5, H = 3 * 3 = 9, F = 3 * 1 = 3.
   The formula is C₅H₉F₃
10. Suppose the theoretical yield in a reaction is 12.45 g of product, but only 10.89 g are obtained. Calculate the percent yield: %Yield = (10.89 g / 12.45 g) * 100% = 87.47%