In: Chemistry
What volume, in mL, of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO3)2?
2Cu(
NO3)2+4KI-----2CuI+I2+4KNO3
calculate mole of Cu(NO3)2
no. of mole = Molarity volume of solution in liter
no. of mole of Cu(NO3)2 = 0.3842 M 0.04388 L = 0.016858696 mole
According to reaction 2 mole of Cu(NO3)2 react with 4 mole of KI then to react with 0.016858696 mole of
Cu(NO3)2 require = 0.016858696 4 / 2 = 0.033717392 mole of KI
volume of solution in liter = no. of mole / Molarity
volume of KI = 0.033717392 / 0.2089 = 0.1614 liter = 161.4 ml
161.4 ml 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO3)2