Q-9) a 500.0-mL sodium iodine (Nal) solution wasprepared by dissolving 36.7 gram

Question

Q-9) a 500.0-mL sodium iodine (Nal) solution wasprepared by dissolving 36.7 grams of solid Nal in water. chlorinegas was bubbled through a 100-0-mL sample of this solution untilthe reaction was complete. how many grams of moecular iodine wereproduced? Q-11) write the balanced molecular equation for thereaction between CaCl2 (aq) and AgNo3 (aq). number______________________- Q-9) a 500.0-mL sodium iodine (Nal) solution wasprepared by dissolving 36.7 grams of solid Nal in water. chlorinegas was bubbled through a 100-0-mL sample of this solution untilthe reaction was complete. how many grams of moecular iodine wereproduced? Q-11) write the balanced molecular equation for thereaction between CaCl2 (aq) and AgNo3 (aq). number______________________-

Explanation / Answer

36.7 g of NaI in 500 mL water. So in 100 mL of this solution, mass of NaI = 36.7 / 5 = 7.34g When chlorine gas is bubbled thru a NaI solution,       2 NaI  + Cl 2   -------->   2NaCl + I2 So 2 moles of NaI give 1 mole of molecular iodine. Moles of NaI = Mass / molar mass                      = 7.34 g / 149.89 g/mol                     = 0.0489 moles 0.0489 moles of NaI will produce 0.0489 / 2 =0.0244 moles of Iodine. Mass of Iodine = Moles * molar mass                         =0.0244 * 253.8 g/mol                         =6.2 g CaCl2 + 2 AgNO3 ------>   Ca(NO3)2   +   2 AgCl So 2 moles of NaI give 1 mole of molecular iodine. Moles of NaI = Mass / molar mass                      = 7.34 g / 149.89 g/mol                     = 0.0489 moles 0.0489 moles of NaI will produce 0.0489 / 2 =0.0244 moles of Iodine. Mass of Iodine = Moles * molar mass                         =0.0244 * 253.8 g/mol                         =6.2 g CaCl2 + 2 AgNO3 ------>   Ca(NO3)2   +   2 AgCl

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