How many liters of chlorine gas, Cl2​, can be obtained at 40°C and 787 mmHg from 9.41 g

Explanation
For this question, we shall be using the ideal gas law equation, PV = nRT. Where P is in atm, V in liters, R = 0.08206 L-at/ mol-K and temperature in K.

First, we determine the number of moles of Cl2(g) generated. Take note that the mole ratio of Cl2 to HCl based on their coefficients in the equation is 5:16. That is, if there are 16 moles of HCl, there will be 5 moles of Cl2 produced.

therefore, we can say that:

Moles of Cl2 = moles of HCl x mole ratio of Cl2 to HCl

We also take note that mole HCl = mass HCl/MW HCl = 9.41g / (1 + 35.5)g/mol

Substituting these to find mole Cl2, we have:

Moles of Cl2 = (mass HCl/MW HCl)( mole ratio of Cl2/HCl ) and substituting the values,

=[(9.41 g)/ (1 +35.5) g/mol][ 5 moles Cl2/ 16 moles HCl)

= 8. 0565 x 10 -2 moles Cl2

Substituting the given values in the problem and the calculated n to the ideal gas law equation, we find the volume:

(787 mmHg/ (760 mmHg/atm))(V) = ( 8. 0565 x 10 -2 moles Cl2)(0.08206 L-atm/mol-K)[(40 +273.15)K]

V= 1.9993 L or simply 2.00 L

Answer
Hello! The answer is V= 2.00 L