Question 9 Sulfur hesxafuoride ts produced by reacting eleme

Question 9 Sulfur hesxafuoride ts produced by reacting elemental sufir with fluorine gas Sy(s) + 24 Fi(x)-, 8 SF6(g) What is the percent yield if 18 3 g SFa is molated from the reaction of 100 g Se and 300gF? 0a402% O b 458% 47.6% 0d546% 0e610% Moving to another question wil save this response.

Ans. #9. Balanced reaction: 8S₁(s) + 24 F₂(g) -----------> 8 SF₆

According to the stoichiometry of balanced reaction, 8 mol S₁ reacts with 24 mol F₂ to produce 8 mol SF₆.

Theoretical molar ratio of reactants: S₁ : F₂ = 8 : 24 = 1 : 3

# Step 1: Determine limiting reagent:

Moles of S₁ taken = Mass / Molar mass = 10.0 g / (32.066 g/ mol) = 0.3119 mol

Moles of F₂ taken = 30.0 g / (38.0 g/ mol) = 0.7895 mol

Experimental molar ratio of reactants = moles of S₁ : Moles of F₂

= 0.3119 mol / 0.7895 mol

= 1 : 2.53

Comparing theoretical and experimental molar ratios of reactants, the experimental moles of F₂ are less than its theoretical value of 3 mol while keeping that of S constant at 1 mol.

So, F₂ is the limiting reactant.

# Step 2: Calculate theoretical yield:

The formation of product follows the stoichiometry of limiting reagent.

So, following stoichiometry-

Theoretical moles of SF₆ formed = (8 SF₆ / 24 F₂) x Moles of F₂ taken

= (8 SF₆ / 24 F₂) x 0.7895 mol

=0.26317 mol

Theoretical moles of SF₆ formed = Theoretical moles x Molar mass

= 0.26317 mol x (146.0564192 g/ mol)

= 38.438 g

So, theoretical yield = 38.438 g

# Step 3: Calculate % Yield:

% yield = (Actual yield/ Theoretical yield) x 100

= (18.3 g / 38.438 g) x 100

= 47.61 %

So, correct option is- C. 47.6 %

#10. Moles of KHCO₃ = 311 g/ (100.11544 g/ mol) = 3.1064 mol

According to the stoichiometry of balanced reaction, 2 mol KHCO₃ produces 1 mol CO₂.

So,

Moles of CO₂ formed = ½ x Moles of KHCO₃

= ½ x 3.1064 mol

= 1.5532 mol

So, correct option is- a. 1.55 mol

#11. Balanced reaction: 6 Mg(s) + P₄(g) -----------> 2 Mg₃P₂

According to the stoichiometry of balanced reaction, 6 mol Mg reacts with 1 mol P₄ to produce 2 mol Mg₃P₂.

Theoretical molar ratio of reactants: Mg : P₄ = 6 : 1

# Step 1: Determine limiting reagent:

Experimental molar ratio of reactants = moles of Mg : Moles of P₄

= 0.14 mol / 0.020 mol

= 7 : 1

Comparing theoretical and experimental molar ratios of reactants, the experimental moles of Mg are greater than its theoretical value of 6 mol while keeping that of P₄ constant at 1 mol.

So, P₄ is the limiting reactant.

# Step 2: Calculate theoretical yield:

The formation of product follows the stoichiometry of limiting reagent.

So, following stoichiometry-

Theoretical moles of Mg₃P₂ formed = (2 Mg₃P₂ / 1 P₄) x Moles of P₄ taken

= (2 Mg₃P₂ / 1 P₄) x 0.020 mol

= 0.040 mol

So, correct option is- e. 0.040 mol

#12. Balanced reaction: C₂H₅OH(l) + 3 O₂(g) ---------> 2 CO₂(g) + 3 H₂O(g)

Moles of C₂H₅OH taken = 3.4 g / (46.06904 g/ mol) = 0.07380 mol

# Following stoichiometry of balanced reaction-

Required moles of O₂ = (3 O₂ / 1 C₂H₅OH) x Moles of C₂H₅OH taken

= (3 O₂ / 1 C₂H₅OH) x 0.07380 mol

= 0.221 mol

Mass of O₂ required = 0.221 mol x (32.0 g/ mol) = 7.08 g

# So, correct option is - NONE

# Note: To completely burn ethanol, a little of excess (around 20%) O₂ is generally required. So, 8.2 g may be chosen. However, since requirement of excess O₂ is NOT mentioned in the question, the value can theoretically be incorrect.

#13. Following stoichiometry of balanced reaction, 1 mol glucose forms 2 mol ethanol.

So,

Theoretical moles of C₂H₅OH formed = 2 x Moles of glucose = 2 x 1.00 mol = 2.00 mol

Theoretical mass of C₂H₅OH formed = 2.00 mol x (46.06904 g/ mol) = 92.13808 g

# yield of C₂H₅OH = (37.0 g / 92.13808 g) x 100 = 40.15 %

So, correct option is- a. 40.2 %

Note: Yeast is NOT a reactant, so its stoichiometry would not be account as not being a part of balanced reaction.