Question

How many grams of HCl are required to produce 11.2 L of Cl2.The reaction occurs in the presence of excess HCl and at STP

Answer 1

Step-by-step explanation:

Equation of reaction:

2HCl → H₂ + Cl₂

From the equation of reaction,

From ideal gas equation,

PV = nRT

P = pressure of the ideal gas

V = volume of the gas

n = number of moles

R = ideal gas constant

T = temperature of the gas

At STP,

P = 1.0atm

T = 273.15K

R = 0.082J/mol.K

PV = nRT

1 .0 × 11.2 = n × 0.082 × 273.15

11.2 = 22.398n

n = 11.2 / 22.398

n = 0.50mole

From the equation of reaction,

2 moles of HCl = 1 mole of Cl₂

X moles HCl = 0.5mole

X = (2 × 0.5) / 1

X = 1 mole of HCL

Number of mole = mass / molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

Mass = number of mole × molar mass

Mass = 1 × 36.5

Mass = 36.5g

36.5g of HCl is required to to produce 11.2L of Cl₂

Answer 2

`m_(HCl)=36.5gHCl`

Step-by-step explanation:

Hello,

In this case, given the reaction:

`2HCl(g)\rightarrow H_2(g)+Cl_2(g)`

We first compute the moles of produced chlorine by knowing 11.2 L were produced at STP conditions, that is 273 K and 1 atm:

`PV_(Cl_2)=n_(Cl_2)RT\\\\n_(Cl_2)=(1atm*11.2L)/(0.082(atm*L)/(mol*K)*273K) =0.5molCl_2`

Hence, since molar mass of hydrochloric acid is 36.45 g/mol and it is in a 2:1 molar ratio with chlorine, we compute the required grams by stoichiometry:

`m_(HCl)=0.5molCl_2*(2molHCl)/(1molCl_2) *(36.45gHCl)/(1molHCl) \\\\m_(HCl)=36.5gHCl`

Best regards.