How many moles of barium hydroxide, #Ba(OH)_2# would be required to react with 117g hydrogen bromide, #HBr# in the reaction #2HBr + Ba(OH)_2 -> BaBr_2 + 2H_2O#?

Answer 1

Approx. 0.725 moles of #Ba(OH)_2#?

Moles of #HBr#? #=# #(117*g)/(80.9*g*mol^-1)# #=# #1.45*mol#.

The stoichiometry of the reaction explicitly tells us that each mole, each equivalent, of barium hydroxide requires 2 mole, 2 equivalents, of hydrogen bromide.

Thus #0.725# #mol# barium hydroxide mole are required.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Owen Ambrose

Calculate the moles of HBr using its molar mass (80.91 g/mol). Identify the mole ratio between HBr and Ba(OH)₂ in the balanced equation (2:1). Determine the moles of Ba(OH)₂ required by applying the mole ratio.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How many moles of barium hydroxide, #Ba(OH)_2# would be required to react with 117g hydrogen bromide, #HBr# in the reaction #2HBr + Ba(OH)_2 -&gt; BaBr_2 + 2H_2O#?

How many moles of barium hydroxide, #Ba(OH)_2# would be required to react with 117g hydrogen bromide, #HBr# in the reaction #2HBr + Ba(OH)_2 -> BaBr_2 + 2H_2O#?