Question

How many moles of Ca[OH]2 are present in 80.0 g of Ca[OH]2?​

Answer 1

1.08mol

Step-by-step explanation:

moles = reacting mass/ molecular weight

reacting mass = 80.0g molecular weight of Ca[OH]2= 40 + 2(16 +1) = 74g/mol

mole = 80.0/74 = 1.08mol