Question

1. What Volume of HCl is generated if 3.44 g of Cl2 are reacted at STP? 2. What volume does 4.87 mol of Kr have at STP? 3. What pressure of HCl is generated if 3.44 g of Cl2 are reacted in 4.55 L at 455 K?

Answer 1

Answer:

1. 2.17 dm3 VOLUME OF HCl IS PRODUCED WHEN 3.44 g OF Cl2 REACT AT STP

2. 4.87 MOLE OF Kr AT STP CONTAINS 109.088 dm3 .

3. THE PRESSURE WHEN 3.44 g OF Cl2 are reacted at 4.55 L AT 455 K IS 0.77 atm

Explanation:

1 Volume of HCl if 3.44 g of Cl2 are reacted at STP?

Equation for the reaction:

H2 + Cl2 ---------> 2HCl

1 mole of Cl2 reacts to form 2 mole of HCl

At STP, 1 mole of a gas is equal to the molar mass of the gas sample

35.5 * 2 g of Cl2 reacts to form 2 mole of HCl

3.44 g of Cl2 will react to form  ( 3.44 * 2 / 71 ) mole of HCl

= 0.0969 mole of HCl

1 mole of HCl = 22.4 dm3

0.0969 mole of HCl = ( 22.4 * 0.0969 / 1)

= 2.17056 dm3

The volume of HCl is 2.17 dm3 when 3.44 g of Cl2 are reacted at STP.

2. What volume does 4.87 mol of Kr have at STP?

1 mole of a substance is 22.4 dm3 of the sample

1 mole of Kr = 22.4 dm3

4.87 mole of  Kr = 4.87 * 22.4

= 109.088 dm3

4.87 mole of Kr at STP contains 109.088 dm3 volume

3. Whta pressure of HCl is generated if 3.44 g of Cl2 are reacted at 4.55 L at 455 K

Using the formula:

PV = nRT

V = 4.55 L

R = 0.082 L atm/ mol K

T = 455 K

m = 3.44 g

n = mass / molar mass

molar mass = ( 1 + 35.5) = 36.5 g/mol

n = 3.44 g / 36.5 g/mol

n = 0.094 mole

P = nRT / V

P = 0.094 * 0.082 * 455 / 4.55

P = 3.50714 / 4.55

P = 0.7708 atm

The pressure of HCl if 3.44 g of Cl2 are reacted at 4.55 L and 455 K is 0.7708 atm.