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.
1. The volume of HCl generated is 2.15 L
2. The volume of 4.87 moles of Kr is 109.088 L
3. The pressure of HCl generated is 0.788 atm
1. Determination of the volume of HCl generated
Mass of Cl₂ = 3.44 g
Molar mass of Cl₂ = 2 × 35.5 = 71 g/mol
Mole = mass / molar mass
Mole of Cl₂ = 3.44 / 71
H₂ + Cl₂ —> 2HCl
From the balanced equation above,
1 mole of Cl₂ reacted to produce 2 moles of HCl.
Therefore,
0.048 mole of Cl₂ will react to produce = 0.048 × 2 = 0.096 mole of HCl
At standard temperature and pressure (STP),
1 mole of HCl = 22.4 L
Therefore,
0.096 mole of HCl = 0.096 × 22.4 = 2.15 L
Thus, the volume of HCl generated at STP is 2.15 L
2. Determination of the volume of 4 moles of Kr at STP
At standard temperature and pressure (STP),
1 mole of Kr = 22.4 L
Therefore,
4.87 moles of Kr = 4.87 × 22.4 = 109.088 L
Thus, 4.87 moles of Kr has a volume of 109.088 L at STP
3. Determination of the pressure of HCl generated.
Number of mole (n) of HCl obtained = 0.096 mole
Volume (V) = 4.55 L
Temperature (T) = 455 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
PV = nRT
P × 4.55 = 0.096 × 0.0821 × 455
P × 4.55 = 3.586128
Divide both side by 4.55
P = 3.586128 / 4.55
Thus, the of HCl generated is 0.788 atm
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