Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

What is the percent yield of either if 1.11L(d=0.7134 g/mL) is isolated from the reaction of 1.500L of C2H5OH

Sagot :

Answer:

Percent yield of ether = 83.18%

Explanation:

Equation of the synthesis reaction of ether from ethanol is given as follows :

2 C₂H₅OH -----> C₂H₅OC₂H₅ + H₂0

Density = mass / volume

Therefore mass = density × volume

Density of ether = 0.7134 g/mL; volume of ether produced = 1.11 L = 1110 mL; Molar mass of ether = 74 g/mol

Mass of ether produced = 0.7134 g/mL × 1110 mL = 791.874 g

Density of ethanol = 0.789 g/mL; volume of ethanol = 1.50 L = 1500 mL; molar mass of ethanol = 46 g/mol

Mass of ethanol reacted = 0.789 g/mL × 1500 mL = 1183.5 g

From the equation of reaction, 2 mole of ethanol produces 1 mole of ether

Mass of 2 moles of ethanol = 2 × 46 = 92 g

Therefore, 92 g of ethanol produces 74 g of ether

1183.5 g of ethanol will produce 1183.5 × 74/92 grams of ether = 951.945 g of ether

Percent yield of ether = actual yield/theoretical yield × 100%

Actual yield of ether = 791.874 g; theoretical yield of ether = 951.945 g

Percent yield of ether = (791.874 g × 951.945 g) × 100 %

Percent yield of ether = 83.18%

We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.