How many grams of HCl are formed if 60.0 g of BCl3 reacts with excess H2O according to the following balanced reaction. BCl3 (g) + 3 H2O (l) → H3BO3 (s) + 3 HCl (g)

Question

Please show work and answer!

Answer 1

BCl₃ (g) + 3H₂O (l) → H₃BO₃ (s) + 3HCl (g) .. balanced equation

molar mas BCl₃ = 117 g / mole

molar mass HCl = 36.5 g / mole

Now we can use the stoichiometry of the balanced equation and the molar masses to answer the question:

60.0g BCl₃ x 1 mol / 117 g x 3 mols HCl / mol BCl₃ x 36.5 g HCl / mol = 56.2 g HCl (3 sig. figs.)