Please help me solve this question
Dear student.
i) 500 cm3 of 0.200 M NaCl No of moles of NaCl = 0.2 x 500 = 100
ii) 100cm3 of 0.500M AgNO3 solution No of moles of AgNO3 = 100 x 0.5 = 50
Reaction involved : AgNO3 + NaCl → AgCl + NaNO3
Part : I Thus, 100 moles of NaCl is added to 50 moles of AgNO3 to produce 50 moles of AgCl as precipitate.
Part : II Weight of AgCl = 50 x molar mass of AgCl = 50 x 143 = 7150 g
Part : III Limiting reagent :
AgNO3 as it will be consumed first due to availability of less no. of moles.
Hope it is clear.
Regards
i) 500 cm3 of 0.200 M NaCl No of moles of NaCl = 0.2 x 500 = 100
ii) 100cm3 of 0.500M AgNO3 solution No of moles of AgNO3 = 100 x 0.5 = 50
Reaction involved : AgNO3 + NaCl → AgCl + NaNO3
Part : I Thus, 100 moles of NaCl is added to 50 moles of AgNO3 to produce 50 moles of AgCl as precipitate.
Part : II Weight of AgCl = 50 x molar mass of AgCl = 50 x 143 = 7150 g
Part : III Limiting reagent :
AgNO3 as it will be consumed first due to availability of less no. of moles.
Hope it is clear.
Regards