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.
