Question

Decane, CiH, undergoes a combustion reaction. Once balanced the coefficient for oxygen is a. 2 b. 10 c. 20 d. 22 e. 31 4. Aluminum metal reacts with aqueous hydrochloric acid to yield hydrogen gas and aqueous aluminum chloride. If 35.9 g of aluminum react with an excess of hydrochloric acid, how many g of hydrogen gas will result? a. 4.91 g b. 3.02 g c. 2.01 g d. 1.70 g e. 4.02 g 3. A 235 g sample of strontium chloride is allowed to react with 99.9 g of sodium phosphate. Strontium chloride and sodium phosphate are soluble in water. Strontium phosphate is insoluble in water. Determine the theoretical yield, in g, of strontium phosphate. a. 672g b. 224 g c. 399 g d. 138 g e. 276 g 4. The volume of a sample of gas at 752 mm Hg is increased from 20.0 L to 22.3 L. Assume constant temperature and number of moles. The final pressure is a. 776 mm Hg b. 711 mm Hg c. 801 mm Hg d. 674 mm Hg e. 838 mm Hg 5. The volume of a gas at 40.00 °C is decreased from 70.00 L to 65.00 L. Assume constant pressure and number of moles. Calculate the final temperature of the gas. a. 290.64 °C b. 64.09 °C c. 313.15 °C d. 337.24°C e. 17.63 °C 6. How many moles of oxygen gas occupy 932 mL of space at 20.0 °C and 755 mm Hg? a. 38.5 moles b. 26.0 moles c. 0.0385 moles d. 29.3 moles e. 0.00456 moles

Answer 1

1. The equation for the combustion of decane is C10H22 + O2 ---------> CO2 + H20 Balance Carbon : C20H22 + 02 ---------> 10 CO2 + H20 Balance Hydrogen: Balance Oxygen: C10H22 + O2 ---------> 10 CO2 + 11 H20 C10H22 + 15.5 02 ---------> 10 CO2 + 11 H20 2 C20H22 + 31 02 ---------> 20 CO2 + 22 H20 Multiply by 2: The coefficient of oxygen = 31 Option E is the right answer. The balanced equation for the reaction is 2 AL + 6 HCI --------> 2 AlCl3 + 3 H2 (g) Number of moles of Al = mass/mol.wt. = 35.9 g/(26.98 g/mol) = 1.330615 mol According to balanced equation 2 moles of Al produces 3 moles of H2 So, 1.330615 mol of Al would produce 1.330615 mol * 3/2 = 1.995923 mol of H2 Mass of H2 = moles * mol.wt. = 1.995923 mol * 2.016 g/mol = 4.02 g Option E is the right answer.

3. The balanced equation is 3 SrCl2 + 2 Na3PO4 ---------> Sr3(PO4)2 (s) + 6 Naci Number of moles of Srcl, = mass/mol.wt. = 235 g/(158.53 g/mol) = 1.482369 mol Number of moles of Na3PO4 = mass/mol.wt. = 99.9 g/(163.94 g/mol) = 0.609369 mol According to balanced equation, 2 moles of Na3PO4 reacts with 3 moles of SrCl2 So, 0.609369 mol of Na3PO, would react with 0.609369 mol * 3/2 = 0.914054 mol of Srci, But, the number of moles of Srcl, present is more than required Hence Srcl, is the excess reagent. According to balanced equation, 2 moles of Na3PO, reacts with 1 moles of Sr3(POA) So, 0.609369 mol of Na3PO, would react with 0.609369 mol * 1/2 = 0.304685 mol of Sr3(PO4)2 Mass of Srz(POA)= moles * mol.wt. = 0.304685 mol * 452.80 g/mol = 138 g Option B is the right answer.

4. Initial pressure of gas = P, = 752 mm Hg Initial volume of the gas = V1 = 20.0 L Final volume of the gas = V2 = 22.3 L Final pressure of the gas = P, = ? According to Boyle's Law, At constant temperature and amount of gas, P.V, = P, V2 P.-PV_ 752 mm Hgx 20.0 L =674.44 mm Hg = 674 mm Hg 22.3 L Option D is the right answer.