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?

Question

Olivia Anton · Accepted Answer

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

#1/2H_2+1/2Br_2 rarr HBr# The formula for "moles of dihydrogen" is (0.92*g)/(2.016*g*mol^-1) = 0.456*mol. "Moles of dibromine" are equal to (45.2-g)/(159.8-g*mol^-1) = 0.283-mol. It is evident that there is a stoichiometric deficiency of halogen present. The quotient must be calculated to determine the yield. #((3.45*g)/(80.91*g*mol^-1))/(1/2xx0.283*mol)xx100%=??#

Peyton Adams · Answer

undefined To calculate the theoretical yield of hydrogen bromide (HBr), first, determine the molar masses of hydrogen (H) and bromine (Br). Then, calculate the number of moles of each reactant using their given masses. The balanced chemical equation for the reaction between hydrogen and bromine to form hydrogen bromide is:

$$ \text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr} $$

Using stoichiometry, determine the limiting reactant. Once the limiting reactant is identified, use its number of moles to calculate the theoretical yield of hydrogen bromide.