Question

Many metals react with halogens to give metal halides. For example, 2Al(s) + 3Cl₂(g) --> 2AlCl₃(s) The above reaction produced 65.3 g of AlCl₃. If you begin with 13.5 g of aluminum, what is the percent yield?

Answer 1

mole = given mass/molar mass

molar mass of Al = 27 g/mol

number of mole Al = 13.5/27

number of mole of Al=0.5

according to balanced chemical equation 2 mole of Al give 2 mole of AlCl3

0.5 mole of Al give 0.5 mole of AlCl3

molar mass of AlCl3= 133.3 g/mol

mass of AlCl3 produce (therotecally)= mole * molar mass

mass of AlCl3 produce(theoretically)= (0.5*133.3) g

mass of AlCl3 produce(theoretically) = 66.6 g

% yield of AlCl3 = (mass of AlCl3 produceexperementally/mass of AlCl3 produce therotically)*100

% yield of AlCl3 =(65.3/66.6)*100

% yield of AlCl3 = 98%

answered by: anonymous