Question

When CS₂(g) (6.828 mol) and 27.31 mol of H₂(g) in a 360.0 L reaction vessel at 382.0 °C are allowed to come to equilibrium the mixture contains 52.58 grams of CH₄(g). What concentration (mol/L) of H₂(g) reacted?

  CS₂(g)+4H₂(g) = CH₄(g)+2H₂S(g)

When SbCl₃(g) (4.334 mol) and 307.3 grams of Cl₂(g) in a 420.0 L reaction vessel at 716.0 °C are allowed to come to equilibrium the mixture contains 0.007729 mol/L of SbCl₅(g). What is the equilibrium concentration (mol/L) of SbCl₃(g)?

SbCl₃(g)+Cl₂(g) = SbCl₅(g)

When H₂(g) (17.50 grams) and 0.02800 mol/L of D₂(g) in a 310.0 L reaction vessel at 757.0 K are allowed to come to equilibrium the mixture contains 6.402 mol of HD(g). What is the equilibrium concentration (mol/L) of H₂(g)?

  H₂(g)+D₂(g) = 2HD(g)

When a sample of (CH₃CO₂H)₂(g) (493.2 grams) is placed in 390.0 L reaction vessel at 560.0 °C and allowed to come to equilibrium the mixture contains 0.01560 mol/L of CH₃CO₂H(g). What concentration (mol/L) of (CH₃CO₂H)₂(g) reacted?

(CH₃CO₂H)₂(g) = 2CH₃CO₂H(g)

When a sample of NO₂(g) (1.005 mol) is placed in 490.0 L reaction vessel at 716.0 °C and allowed to come to equilibrium the mixture contains 26.17 grams of N₂O₄(g). What is the equilibrium concentration (mol/L) of NO₂(g)?

2NO₂(g) = N₂O₄(g)

When a sample of ClF(g) (7.263 mol) is placed in 56.00 L reaction vessel at 695.0 °C and allowed to come to equilibrium the mixture contains 60.06 grams of F2(g). What is the concentration (mol/L) of F2(g)?   2ClF(g) = F2(g)+Cl2(g)

  

Answer 1

1. CS₂(g)+4H₂(g) = CH₄(g)+2H₂S(g)

To form 1 mol of methane, 2 mol of hydrogen gas is required and to form 1 mol of H₂S 1 mol of hydrogen gas is required.

According to the question, 52.58 g, i.e., (52.58/16)=3.28 mol of methane is formed. and (2 × 3.28)= 6.56 mol of H₂S gas is formed.

Now, (2 × 3.28) mol of hydrogen is reacted to form 3.28 mol of methane and 6.56 mol of hydrogen gas is required to form 6.56 mol of H₂S.

A total of ( 6.56 +6.56) mol or 13.12 mol hydrogen gas is reacted.

Density of hydrogen gas reacted = 13.12/ 360 = 0.036 mol/L

2. SbCl₃(g)+Cl₂(g) = SbCl₅(g)

Total 0.007729 * 420 = 3.25 mol of SbCl₅(g) was formed. Amount of SbCl₃(g) at equilibrium = 4.334 - 3.25 = 1.084 mol ....[ as 3.25 mol SbCl₃(g) is required to form 3.25 mol of SbCl₅(g)]

Equilibrium concentration of SbCl₃(g) = 1.084/ 420 = .00258 mol/L

3. H₂(g)+D₂(g) = 2HD(g)

To form 1 mol HD, 1/2 mol of H₂ gas is required. So, 6.402 mol HD = 6.402/2 mol H₂

3.2 mol H₂ reacted.

Amount of H₂ before reaction = 17.5/2 = 8.75 mol.

Conc of H₂ at equilibrium = 8.75-3.2/310 = 0.0179mol/L

4. (CH₃CO₂H)₂(g) = 2CH₃CO₂H(g)

1 mol of CH₃CO₂H)₂ gives 2 mol of CH₃CO₂H(g).

Amount of (CH₃CO₂H)₂ before reaction = 493.2/ 120 = 4.11 mol

0.01560 mol/L of CH₃CO₂H(g) in the vessel means 0.01560 * 390 = 6.084 mol of CH₃CO₂H(g)

Amount of (CH₃CO₂H)₂ reacted = 6.084/2 = 3.042 mol = 3.042/390 = 0.0078 mol/L

5. 2NO₂(g) = N₂O₄(g)

Do similarly as above.. (1 mol NO₂ gives 1 mol of N₂O₄(g) )

Ans: 0.00147mol/L

6. 60.06 grams of F₂(g) = 60.06/ 38 = 1.58 mol = 1.58/56 = 0.028 mol/L