Question

3.45 g of hydrogen bromide (HBr) were prepared by reacting 0.920 g of hydrogen and 45.2 g of bromine. How do you calculate the theoretical yield?

Answer 1

Well, write the stoichiometric equation to inform your reasoning...and get approx. `30%`…….

Explanation:

`1/2H_2+1/2Br_2 rarr HBr`

`"Moles of dihydrogen"=(0.92*g)/(2.016*g*mol^-1)=0.456*mol`

`"Moles of dibromine"=(45.2*g)/(159.8*g*mol^-1)=0.283*mol`

Clearly, the halogen in present in stoichiometric deficiency, agreed? For the yield we need to take the quotient...

`((3.45*g)/(80.91*g*mol^-1))/(1/2xx0.283*mol)xx100%=??`