Question

What is the percent yield of AlBr3 if 120.7 grams of AlBr3 is produced in when 12.7 g of aluminum reacts? Round to two decimal places and do not include a % sign in your answer.


2Al + 3Br2 → 2AlBr3

Answer 1

2Al + 3Br₂$\to$ 2AlBr₃

theoretical yield = (m(Al)*2*M(AlBr₃))/2*M(Al)=$(12.7g*2*266.69g/mol)/(2∗26.98g/mol)=125.536g$

%yield=actual yield/theoretical yield*100%=$120.7g/125.536g*100 =96.15$