Question

How many molecules of HCl are formed when 90.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 

_______  ICl₃ + _______ H₂O → _______ ICI + _______ HIO₃+ _______ HCI 

_______ molecules HCI

Answer 1

The Balanced equation for the given reaction is :

2 ICl₃ +3 H₂O → ICl + HIO₃ + 5 HCl

So from the above equation it is observed that

3 moles of H₂O = 5 moles of HCl

(1 mole of H₂O = 18 g and 1 mol of HCl = 6.022 x 10²³ molecules)

⟹ 3 x 18 g of  H₂O = 5 moles of HCl

⟹ (3 x 18) g of  H₂O = 5 x 6.022 x 10²³ molecules of HCl

⟹ 1 g of  H₂O = (5 x 6.022 x 10²³) / (3 x 18) molecules of HCl

⟹ 1 g of  H₂O = 5.58 x 10²² molecules of HCl

Therefore , 90.0 g g of  H₂O = 90.0 x 5.58 x 10²² molecules of HCl = 5.02 x 10²⁴ molecules of HCl

Hence , 90.0 g of H₂O produces 5.02 x 10²⁴ molecules of HCl