If 5.86g of glucose are reacted and 1.44L of CO2 gas are collected at 293K and .984atm. What is the percent yield of the reaction?

C6H1206 -> 2C2H5OH + 2CO2

Here is an example stoichiometry problem that shows percent yield at the end. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

I'm really confused.. I got .92 % is that correct?

To determine the percent yield of the reaction, we need to compare the actual yield to the theoretical yield. The theoretical yield can be calculated based on the given mass of glucose reacted.

Step 1: Calculate the molar mass of glucose (C6H12O6).
The molar mass of glucose can be calculated by adding up the atomic masses of each element present in the compound.
C: 6 atoms x 12.01 g/mol = 72.06 g/mol
H: 12 atoms x 1.01 g/mol = 12.12 g/mol
O: 6 atoms x 16.00 g/mol = 96.00 g/mol
Total molar mass = 72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol

Step 2: Convert the mass of glucose to moles.
To convert the mass of glucose (5.86 g) to moles, divide the given mass by the molar mass.
Moles of glucose = 5.86 g / 180.18 g/mol = 0.0325 mol

Step 3: Use stoichiometry to determine the expected moles of CO2 produced.
From the balanced chemical equation, we know that 1 mol of glucose reacts to produce 2 mol of CO2.
Moles of CO2 = 2 mol of CO2/mol of glucose x 0.0325 mol = 0.065 mol

Step 4: Convert moles of CO2 to liters using the ideal gas law.
At STP (standard temperature and pressure), 1 mol of ideal gas occupies 22.4 L.
The given reaction conditions are not at STP, so we need to use the ideal gas law equation:
PV = nRT

V = (nRT) / P
V = (0.065 mol x 0.0821 L·atm/(mol·K) x 293 K) / 0.984 atm
V = 1.56 L

Step 5: Calculate the percent yield.
The percent yield is calculated by dividing the actual yield (1.44 L of CO2 gas) by the theoretical yield (1.56 L of CO2 gas) and multiplying by 100%.
Percent yield = (1.44 L / 1.56 L) x 100% ≈ 92.3%

Therefore, the percent yield of the reaction is approximately 92.3%.

To calculate the percent yield of a reaction, you need to know the actual yield (the amount of product obtained experimentally) and the theoretical yield (the amount of product that would be obtained if the reaction went to completion based on stoichiometry).

In this case, we are given the mass of glucose reacted (5.86g) and the volume of CO2 gas collected (1.44L). However, we need to find the actual yield, which is the amount of ethanol produced in the reaction.

To find the actual yield of ethanol, we need to use stoichiometry. The balanced chemical equation tells us that 1 mole of glucose (C6H12O6) reacts to form 2 moles of ethanol (C2H5OH) and 2 moles of CO2.

To determine the number of moles of glucose, we can use its molar mass. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.

moles of glucose = mass / molar mass
moles of glucose = 5.86g / 180.16 g/mol

Next, we use the stoichiometry of the reaction to calculate the moles of ethanol produced. Since 1 mole of glucose gives 2 moles of ethanol, we can use the mole ratio:

moles of ethanol = 2 moles of ethanol / 1 mole of glucose * moles of glucose

Now we have the moles of ethanol produced. To convert this to mass, we use the molar mass of ethanol, which is approximately 46.07 g/mol.

mass of ethanol = moles of ethanol * molar mass
mass of ethanol = moles of ethanol * 46.07 g/mol

Now we have the actual yield of ethanol in grams.

To calculate the theoretical yield of ethanol, we would consider the limiting reagent in the reaction. However, we do not have the amount of the limiting reagent given in the question. So, we cannot determine the theoretical yield from the given data.

Lastly, we can calculate the percent yield using the formula:

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

Since we do not have the value of the theoretical yield, we cannot calculate the percent yield.