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

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%

i think it's c. i'm not sure but i think so.