Question

2:06 LTED Question 10 of 20 Submit How many liters of a 0.209 MKI solution is needed to completely react with 2.43 g of Cu(NO3)2 according to the balanced chemical reaction: 2 2 Cu(NO3)2(aq) + Cul(aq) + 12(s) + 4 4 Kl(aq) → KNO3(aq) STARTING AMOUNT ADD FACTOR ANSWER RESET 2.43 2 0.124 6.022 x 1023 187.57 0.00542 0.0310 0.0620 0.209 166 4 Tap here or pull up for additional resources

Answer 1

Answer: Given that, Molarity of KI = 0.209 M Mass of Cu CNO3)2 = 2.43 g Molari mass of Cu CNO3) = 187.56 g/mol So, No. of moles of Cu(NO3)2 - given mass - 2.43 g - Molat mass moles of Cu (NO3)2 = 0.013 males The balance reaction is 2 Cu(NO3), caq) + 4 KT cag) → 2 Cut cap) +T26 + 4 kNO, (01) 187.56 g/mol 2 male of eu (NO3), react with 4 mL of KI So, 0.013 moles of cu(NO3)2 will react with D = 24 x 0.013 molas = 0.026 moles of kI Now, Volume of KI Volume of KI = us Monty 126 no. of males of KI Molarity of KS 0.026 o dog = 0.1244 L V