Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction

74%

Explanation:

Step 1: Write the balanced equation

2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.

The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.

Since EMR > TMR, the limiting reactant is O₂.

Step 3: Calculate the theoretical yield of H₂O

The theoretical mass ratio of O₂ to H₂O 544:180.

199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O

Step 4: Calculate the percent yield of H₂O

%yield = (experimental yield/theoretical yield) × 100%

%yield = (49 g/65.8 g) × 100% = 74%

Percentage yield of H₂O = 74.24%

Explanation:

The balanced equation for the reaction is given below:

2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O

Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:

Molar mass of C₆H₁₀ = 82 g/mol

Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 17 × 32 = 544 g

Molar mass of H₂O = 18 g/mol

Mass of H₂O from the balanced equation = 10 × 18 = 180 g

SUMMARY:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂.

Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.

Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.

Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:

From the balanced equation above,

544 g of O₂ reacted to produce 180 g of H₂O.

Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.

Thus, the theoretical yield of H₂O is 66 g.

Finally, we shall determine the percentage yield. This can be obtained as follow:

Actual yield of H₂O = 49 g

Theoretical yield of H₂O = 66 g

Percentage yield of H₂O =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of H₂O = 49/66 × 100

Percentage yield of H₂O = 74.24%