Calculate the percent yield if 52.9 g of ethanol reacts to produce 26.7 g of ether.
2ch3ch2oh(l) → ch3ch2och2ch3(l) + h2o(l)

The % yield is 62.7 %

Explanation:

Step 1: the balanced equation

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

Step 2: Data given

mass of ethanol = 52.9 grams

molar mass of ethanol = 46.07 g/mol

mass of ether = 26.7 grams

Molar mass of ether = 74.12 g/mol

Step 3: Calculate moles of ethanol

Number of moles = Mass / molar mass

Moles of ethanol = 52.9 grams / 46.07 g/mol

Moles of ethanol = 1.15 moles

Step 4: Calculate moles of ether

For 2 moles of ethanol consumed, we have 1 mole of ether produced

For 1.15 moles of ethanol consumed, we have 1.15/2 = 0.575 moles of ether produced

Step 5: Calculate mass of ether

Mass = moles * Molar mass

Mass ether = 0.575 moles * 74.12 g/mol = 42.62 grams = theoretical yield

Step 6: Calculate % yield

% yield = (26.7 / 42.62) * 100 = 62.7%

The % yield is 62.7 %

the answer is learned

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !