Question

Gaseous butane (C4H10) reacts with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O) as follows:

2C4H10(g)+13O2(g)=8CO2(g)+10H2O(g)
If 4.10 g of water is produced from the reaction of 5.23 g of butane and 11 g of oxygen gas, what is the percent yield of water?

Answer 1

To find the percent yield of water, we need to calculate the theoretical yield of water and actual yield of water, and then use the formula: Percent Yield = (Actual Yield / Theoretical Yield) * 100%. The percent yield in this reaction is approximately 25.56%.

Step-by-step explanation:

The given reaction is:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

To find the percent yield of water, we need to calculate the theoretical yield of water and actual yield of water, and then use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

First, calculae the molar mass of butane (C4H10):

(4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol

Next, calculate the molar mass of water (H2O):

(2 * 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol

Now, calculate the number of moles of butane: 5.23 g / 58.12 g/mol = 0.09 mol

Calculate the number of moles of oxygen:

11 g / 32.00 g/mol = 0.34 mol

Since the coefficient of water in the balanced equation is 10, the theoretical yield of water can be calculated as:

0.09 mol * 10 = 0.9 mol

To find the actual yield of water, convert the given mass of water to moles:

4.10 g / 18.02 g/mol = 0.23 mol

Finally, plug the values into the percent yield formula:

Percent Yield = (0.23 mol / 0.9 mol) * 100% = 25.56%

Therefore, the percent yield of water in this reaction is approximately 25.56%.