how many moles of barium hydroxide, ba(oh)2, would be required to react with 117 g hydrogen bromide, hbr…

how many moles of barium hydroxide, ba(oh)2, would be required to react with 117 g hydrogen bromide, hbr? 2hbr + ba(oh)2 → babr2 + 2h2o ? mol ba(oh)2

how many moles of barium hydroxide, ba(oh)2, would be required to react with 117 g hydrogen bromide, hbr? 2hbr + ba(oh)2 → babr2 + 2h2o ? mol ba(oh)2

Answer

Explanation:

Step1: Calculate moles of HBr

The molar - mass of HBr is (M_{HBr}=1 + 79.904=80.904\ g/mol). Using the formula (n=\frac{m}{M}), where (m = 117\ g) and (M = 80.904\ g/mol), we have (n_{HBr}=\frac{117\ g}{80.904\ g/mol}\approx1.446\ mol).

Step2: Use mole - ratio from the balanced equation

From the balanced chemical equation (2HBr+Ba(OH)2\rightarrow BaBr_2 + 2H_2O), the mole - ratio of (HBr) to (Ba(OH)2) is (2:1). Let the moles of (Ba(OH)2) be (n{Ba(OH)2}). Then (n{Ba(OH)2}=\frac{n{HBr}}{2}). Substituting (n{HBr}=1.446\ mol) into the equation, we get (n{Ba(OH)_2}=\frac{1.446\ mol}{2}=0.723\ mol).

Answer:

(0.723)