In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide

C6H1206 -----> 2C2H5OH + 2CO2

If 5.97 g of clucose are reacted and 1.44 L of CO2 gas are collected at 293 K and .984 atm, what is the percent yield of the reaction?

Convert 5.97 g glucose to mols.

Convert mols Glucose to mols CO2 using the coefficients in the balanced equation.
Convert mols CO2 at STP to volume (22.4 L/mol) CO2 at STP. This is the theoretical yield.
Now use PV = nRT to convert 1.44 L CO2 at the cnoditions listed to mols of CO2, then convert to volume CO2 at STP (again, 22.4 L/mol at STP conditions). This is the actual amount produced.
%yield = [actual yield/theoretical yield]*100 = ??
Post your work if you get stuck.

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Well, it seems we have a ferment-tation question here! Let's do some number crunching, shall we?

First, we need to convert our grams of glucose to moles. The molar mass of glucose (C6H12O6) is around 180.156 g/mol. So, dividing 5.97 g by 180.156 g/mol, we get approximately 0.0331 moles of glucose.

Next, we look at the balanced equation: 1 mole of glucose produces 2 moles of CO2. So, since we reacted 0.0331 moles of glucose, we should expect to produce 2 * 0.0331 = 0.0662 moles of CO2.

Now, we can use the ideal gas law to find the volume of CO2 at STP (standard temperature and pressure). The ideal gas law formula is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

Converting 293 K to Celsius, we get 20 °C (because we always need to know the cool temperature, right?). Plugging in the numbers, we have (0.984 atm) * (V) = (0.0662 mol) * (0.0821 L·atm/(mol·K)) * (293 K).

Solving for V, we find V ≈ 1.72 L of CO2 at STP.

Now we can calculate the percent yield. The theoretical yield (the maximum amount of product we can get) is 1.72 L, while the actual yield is given as 1.44 L. To find the percent yield, we divide the actual yield by the theoretical yield and multiply by 100.

So, (1.44 L / 1.72 L) * 100 = approximately 83.72%.

Therefore, the percent yield of the reaction is around 83.72%, which means some gas was just having a little too much fun and couldn't be contained in the collector. Remember, in life, it's all about the effort you put in (and a little bit of CO2 escapades)!

Hope that answers your question!

To determine the percent yield of the reaction, we need to compare the actual yield with the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly according to the balanced equation.

1. Calculate the moles of glucose reacted:
To do this, we need to know the molar mass of glucose. The molar mass of glucose (C6H12O6) is:
6 * Atomic mass of carbon (C) + 12 * Atomic mass of hydrogen (H) + 6 * Atomic mass of oxygen (O)

The atomic masses are:
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of hydrogen (H) = 1.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol

Calculating the molar mass of glucose:
Molar mass of glucose = 6 * 12.01 g/mol + 12 * 1.01 g/mol + 6 * 16.00 g/mol

Now, we can calculate the moles of glucose reacted:
moles of glucose = mass of glucose / molar mass of glucose

Substitute the given values:
moles of glucose = 5.97 g / (6 * 12.01 g/mol + 12 * 1.01 g/mol + 6 * 16.00 g/mol)

2. Calculate the moles of carbon dioxide produced:
According to the balanced equation, 1 mole of glucose produces 2 moles of carbon dioxide. Therefore,
moles of carbon dioxide = 2 * moles of glucose

3. Calculate the volume of carbon dioxide at STP:
We know that 1 mole of any gas occupies a volume of 22.4 L at STP (standard temperature and pressure). However, the given conditions are not at STP. To convert to STP, we can use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

We rearrange the equation to solve for the moles using the given values:
n = PV / RT

Substitute the values:
moles of carbon dioxide = (0.984 atm) * (1.44 L) / (0.0821 L·atm/(mol·K) * 293 K)

4. Calculate the theoretical yield of carbon dioxide:
Using the moles of carbon dioxide calculated in step 2, we can calculate the mass of carbon dioxide produced. The molar mass of carbon dioxide (CO2) is:
molar mass of carbon dioxide (CO2) = Atomic mass of carbon (C) + 2 * Atomic mass of oxygen (O)

The atomic masses are:
Atomic mass of carbon (C) = 12.01 g/mol
Atomic mass of oxygen (O) = 16.00 g/mol

molar mass of carbon dioxide (CO2) = 12.01 g/mol + 2 * 16.00 g/mol

Now, we can calculate the theoretical mass of carbon dioxide produced:
theoretical mass of carbon dioxide = moles of carbon dioxide * molar mass of carbon dioxide

5. Calculate the percent yield:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

percent yield = (actual yield / theoretical yield) * 100

Substitute the values:
percent yield = (mass of carbon dioxide collected / theoretical mass of carbon dioxide) * 100

By following these steps and substituting the given values into the formulas, you should be able to calculate the percent yield of the reaction.