Question

Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following balanced equation:2KI(aq) + Cl2(

g. → 2KCl(aq) + I2(
g.How many moles of I2 are produced?

Answer 1

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

Step-by-step 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

Answer 2

Answer: 0.3794 moles of Iodine gas are produced.

Step-by-step explanation:

`2KI(aq)+Cl_2( g)\rightarrow 2KCl(aq)+I_2`

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:
`(1)/(22.4 L)* 8.5=0.3794 moles`

0.3794 moles of Iodine gas are produced.