BaO + H20
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\u00d7171.34"
"=1028.04g"
Comments
Leave a comment