Respuesta :
Answer: 0.3794 moles of Iodine gas are produced.
Explanation:
[tex]2KI(aq)+Cl_2( g)\rightarrow 2KCl(aq)+I_2[/tex]
Volume of iodine gas produced at STP =8.5 L
At STP, the 1 mol of gas occupies volume = 22.4 L
So, 8.5 L of volume will be occupied by:[tex]\frac{1}{22.4 L}\times 8.5=0.3794 moles[/tex]
0.3794 moles of Iodine gas are produced.
Ans: 0.38 moles of I2 are produced.
Given:
Volume of I2, V = 8.5 L
At STP condition:
Temperature, T = 273 K
Pressure, P = 1 atm
To determine:
moles of I2 produced
Explanation:
Based on the ideal gas law:
PV = nRT
n = # moles of gas
R = gas constant = 0.0821 L.atm/mol-K
n = PV/RT
= 1 atm*8.5 L/0.0821 L.atm/mol-K*273 K = 0.379 moles
