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

Answer:

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

Answer:

$m_{HCl}=36.5gHCl$

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:

$PV_{Cl_2}=n_{Cl_2}RT\\\\n_{Cl_2}=\frac{1atm*11.2L}{0.082\frac{atm*L}{mol*K}*273K} =0.5molCl_2$

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:

$m_{HCl}=0.5molCl_2*\frac{2molHCl}{1molCl_2} *\frac{36.45gHCl}{1molHCl} \\\\m_{HCl}=36.5gHCl$

Best regards.