Ethanol (c2h5oh) is produced from the fermentation of sucrose in the presence of enzymes.
c12h22o11(aq) + h2o(g) 4 c2h5oh(l) + 4 co2(g)
determine the theoretical yield and the percent yields of ethanol if 735 g sucrose undergoes fermentation and 310.5 g ethanol is obtained.
theoretical
g
percent
%

Explanation:

to calculate theoratical yield...first we must calculate how much Sucrose moles we have..in order to to do so...we must calculate molar mass of sucrose.

molar mass of sucrose is 342.3 g/mol.

now we can calculate how many sucrose moles we have by dividing the mass with molar mass of sucrose

735g/(342.3g/mol)=2.147mol

theoratically...according to stoichiometry..every 1 mole of sucrose yields 4 moles of ethanol...so...

2.147mole yield 2.147*4mol=8.588mol

now we must calculate weight of that much ethanol..molar mass of ethanol is 46.07 g/mol...

so we can multiple moles by molar mass to obtain the weight 8.588mol*46.07g/mol=395.649g

but we only obtained 310.5g...so percentage we have is

78.47%

if there is trouble with molar mases I used....use what you calculated...

If my explanation is good enough...please mark it as brainliest.thanks

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation: