53. A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of

Question

53. A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is The following reaction occurs when the two solutions are mixed. HCI(aq) + NHsag NH Cl(aq). If reacted? 480. J/K. + NH (aq he temperature increase is 2.34°C. Calculate H per mole of HCl and NH A)154 kJ/mol B) 1.96 kJ/mol C)100 kJ/mol D) -1.96 kJ/mol E) -154 kJ/mol 54. The enthalpy change when a strong acid is neutralized by strong base is-56.1 kJ/mol. If 135 mL of 0.450 M HI at 23.15°C is mixed with 145 mL of0.500 M NaOH, also at 23.15 what will the maximum temperature reached by the resulting solution? (Assume that there is no heat loss to the container, that the specific heat of the final solution is 4.18 J/g: C, and that the density of the final solution is that of water. A) 26.06°C B) 29.19°C C) 32.35°C D) 20.24°C E) 36.57°C 55. How many liters of oxygen are needed to exactly react with 19.8 g of methane at STP? CH4(g) + 2 O2(g) CO2(g) + 2 H2O() A) 13.9 L B) 27.8 L C) 55.5 L D) 60.5 L 56. The titration of 80.0 mL of an unknown concentration H3PO4 solution requires 126 mL of 0.218 M KOH solution. What is the concentration of the H3PO4 solution (in M)?

Explanation / Answer

55. CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2 H2O (g)

Moles of CH4 = 19.8 g * 1 mol/ 16.0 g =1.238 mol CH4

Moles of O2 required to react with 1.234 mol CH4 = 1.238 mol CH4 * 2 mol O2/ 1 mol CH4

= 2.476 mol O2

Volume of O2 required to react with 1.234 mol CH4 at STP= 2.476 mol O2 * 22.4 L /1 mol O2

     =55.5 L O2

56. H3PO4 + 3 KOH ------> K3PO4 + 3 H2O (l)

Moles of KOH = 0.218 mol/L * (126 mL * 1 L /1000 mL ) = 0.027468 mol KOH

Moles of H3PO4 = 0.027468 mol KOH * 1 mol H3PO4 / 3 mol KOH =0.009156 mol H3PO4

Molarity of H3PO4 = 0.009156 mol H3PO4 / (80.0 mL*1L/1000mL) = 0.11445 M H3PO4

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