To determine the amount of medicinal ether produced from 50 grams of ethyl alcohol with an 85% yield, we need to know the chemical reaction involved in the production of ethers. In this case, ethyl alcohol (ethanol) can react to form diethyl ether (medicinal ether) when it is treated with an acid.
The balanced chemical reaction for the formation of diethyl ether from ethanol can be represented as follows:
\[ 2 , \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Acid}} \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{H}_2\text{O} \]
From the reaction, we see that 2 moles of ethanol produce 1 mole of diethyl ether.
Step 1: Calculate the number of moles of ethanol (C2H5OH)
The molar mass of ethanol (C2H5OH) is: \[ 2(12.01) + 6(1.008) + 16.00 = 46.07 , \text{g/mol} \]
Calculating the number of moles of ethanol in 50 grams: \[ \text{Moles of ethanol} = \frac{50 , \text{g}}{46.07 , \text{g/mol}} \approx 1.084 , \text{mol} \]
Step 2: Determine the moles of diethyl ether produced
According to the balanced reaction, 2 moles of ethanol yield 1 mole of diethyl ether. Therefore, the moles of diethyl ether produced would be: \[ \text{Moles of diethyl ether} = \frac{1.084 , \text{mol}}{2} \approx 0.542 , \text{mol} \]
Step 3: Calculate the mass of diethyl ether produced
Next, we need to calculate the molar mass of diethyl ether (C2H5OC2H5): \[ 2(12.01) + 10(1.008) + 16.00 = 74.12 , \text{g/mol} \]
Now we can calculate the theoretical mass of diethyl ether: \[ \text{Mass of diethyl ether} = 0.542 , \text{mol} \times 74.12 , \text{g/mol} \approx 40.19 , \text{g} \]
Step 4: Calculate the actual yield based on percent yield
To find the actual yield based on the percent yield: \[ \text{Actual mass} = \text{Theoretical mass} \times \frac{\text{percent yield}}{100} \] \[ \text{Actual mass} = 40.19 , \text{g} \times \frac{85}{100} \approx 34.16 , \text{g} \]
Therefore, the amount of medicinal "ether" produced from 50 grams of ethyl alcohol with an 85% yield is approximately 34.16 grams.
1 answer