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

asked Jul 23, 2021 52.1k views

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Answer:

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

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