# 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?

Jun 16, 2018

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

#### Explanation:

$\frac{1}{2} {H}_{2} + \frac{1}{2} B {r}_{2} \rightarrow H B r$

$\text{Moles of dihydrogen} = \frac{0.92 \cdot g}{2.016 \cdot g \cdot m o {l}^{-} 1} = 0.456 \cdot m o l$

$\text{Moles of dibromine} = \frac{45.2 \cdot g}{159.8 \cdot g \cdot m o {l}^{-} 1} = 0.283 \cdot m o l$

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%=??