Chlorine and hydrogen gas react to form hydrogen chloride as shown in the following reaction: Cl 2 (g)+H 2 (g) 2HCl (g) If a 48.5 L sample of chlorine gas was reacted with excess hydrogen at 450 K and 1.20 atm , how many grams of hydrogen chloride is produced ?

It’s 1.58 moles

Answers

Answer 1

Answer:

H

Explanation:

Answer 2

Approximately 1229.80 grams of hydrogen chloride (HCl) is produced.

We must use the specified number of moles of the limiting reactant, in this case chlorine gas [tex](Cl_2)[/tex], to calculate the mass of hydrogen chloride (HCl) produced during the reaction. We can see from the balanced equation that 1 mole of [tex]Cl_2[/tex] results in the formation of 2 moles of HCl.

Given:

Volume of Cl2 gas = 48.5 L

Temperature (T) = 450 K

Pressure (P) = 1.20 atm

Moles of Cl2 = 1.58 moles

The ideal gas law equation can be used to calculate the amount of HCl produced:

PV = nRT

Where:

P = pressure

V = volume

n = moles

R = ideal gas constant

T = temperature

First, we must use the ideal gas law equation to obtain the amount of HCl in moles:

n(HCl) = (P * V) / (R * T)

Using the known values:

P = 1.20 atmV = 48.5 LR = 0.0821 L·atm/(K·mol)T = 450 K

n(HCl) = (1.20 atm * 48.5 L) / (0.0821 L·atm/(K·mol) * 450 K)

≈ 33.73 moles

To the nearest two decimal places, the mass of HCl is equal to its molar mass times its molar mass, or 33.73 mol times 36.46 g/mol, or 1229.80 g.

Mass of HCl = moles of HCl * molar mass of HCl

= 33.73 moles * 36.46 g/mol

≈ 1229.80 grams (rounded to two decimal places)

Therefore, approximately 1229.80 grams of hydrogen chloride (HCl) is produced.

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