Question

 BaO + H20

Answer 1

Kindly note the question is incomplete

Adjusting the condition of the question

"Given the following reaction

BaO+ H₂ O $\to$ Ba(OH)₂

Determine the number of grams of Ba(OH)₂ formed when 920g of BaO and 300g of H₂O react"

Solution

BaO$=153.33g/mol$

H₂O$=18.01528g/mol$

Ba(OH)₂ $=171.34g/mol$

$920g$ of BaO $=\frac{920}{153.33}=6.0moles$

$300g$ of water $=\frac{300}{18.01528}=16.6525$

Reacting Ratio $=1:1$

$\therefore$ Water is in excess

1 mole of BaO produce 1 mole of Ba(OH)₂

$\therefore$ $6$ moles of Ba(OH)₂ will be produced

Mass of Ba(OH)₂ $= 6×171.34$

$=1028.04g$