CHE 111 General Chemistry 1-SP18 Homework Set 22-25 pts Name: The following home
ID: 1034423 • Letter: C
Question
CHE 111 General Chemistry 1-SP18 Homework Set 22-25 pts Name: The following homework problems are due on Friday Mar 30th, at the end of class. You may work together on these, but make sure that you contribute to the group and can perform these on your own on similar problems that may appear on the test Ch 7 46. (3 pts) H2 is produced by the reaction of 118.5 mL of a 0.8775-M solution of H PO according to the following equation: Determine the number of moles and mass of H2 that are produced in this reaction. 2 Cr(s)+2 H,PO.(aq) 3 Hz(g)+2 CrPO(aq) 50. (3 pts) What mass of silver oxide, AgzO, is required to produce 25.0 g of silver sulfadiazine, AgC10H N4SO2, from the reaction of silver oxide and sulfadiazine? 56. (4 pts) What volume of 0.750 M hydrochloric acid solution can be prepared from the HC produced by the reaction of 25.0 g of NaCI with excess sulfuric acid? NaCI(s)+ H2SO4) HCI(g)+NaHSO (s)Explanation / Answer
46. According to the given reaction, for the consumption of 2 moles of H3PO4, 3 moles of hydrogen is produced.
The no. of moles of H3PO4 in the given problem = 118.5 mL * 0.8775 mmol/mL = 103.98375 mmol ~ 0.104 mol
Therefore, for the consumption of 0.104 moles of H3PO4, the no. of moles of H2 produced = 0.104*3/2 = 0.156 mol
The molar mass of H2 = 2 g/mol
Hence, the mass of H2 = 0.156 mol * 2 g/mol = 0.312 g
50. According to the given reaction, for the production of 2 moles of silver sulfadiazine, 1 mole of silver oxide is required.
The molar mass of silver sulfadiazine = 357.136 g/mol
25 g. of silver sulfadiazine = 25 g/357.136 g mol-1 ~ 0.07 mol
Therefore, the no. of moles of silver oxide required = 0.07*1/2 = 0.035 mol
The molar mass of silver oxide = 231.735 g/mol
Hence, the mass of silver oxide required = 0.035 mol * 231.735 g/mol = 8.111 g
56. According to the given reaction, 1 mole of HCl is produced by the reaction of 1 mole of NaCl.
The molar mass of NaCl = 58.5 g/mol
The no. of moles of NaCl in the given problem = 25 g/58.5 g mol-1 = 0.42735 mol
Therefore, 0.42735 moles of HCl will be produced.
The volume of HCl solution required = 0.42735 mol/0.75 mol L-1
= 0.57 L = 570 mL (since 1 L = 1000 mL)